Gravitation

(a). No

Charges are shielded from electrostatic forces by keeping them inside hollow conductor as no electric field penetrates the conductor and both gravitational forces and electric forces are field forces i.e. A body or charge experience force due to other body because of interaction between its own field and field due to other body or charge. We does not allow external field to interact with field of charge so it does not experience any force but there is no method to shield a body from gravity because gravity is independent of medium and there is no material found which shields gravitational field or does not allow it to pass through it so external gravitational field interacts with gravitational field of body and hence body experience force.

(b). Yes

If the spaceship is large enough then the astronaut will definitely detect the Earth’s gravity as Gravitational Force on spaceship is directly proportional to the mass of spaceship so as mass of bigger spaceship will be large and hence if will experience noticeable amount of force which can be detected, we know gravitational force on a body is given as

Where, F is gravitational force

G is universal gravitational Constant

m₁ is mass of first body which in this case is earth

m₂ is the mass of second body which is spaceship

and r is distance between earth and spaceship

so as mass of spaceship m₂ increases the gravitational force experienced by it increases and hence can be detected i.e. Gravity can be detected

(c). As we know Gravitational force is inversely proportional to the square of the distance

Where FG is gravitational force and r is the distance

Where G is universal gravitational Constant

m₁ is mass of first body which in this case is moon

m₂ is the mass of second body which is Earth or moon

and r is distance between earth and spaceship

so as the distance increases Gravitational force between bodies decreases and sun is at a far more distance than moon but mass of earth is so much greater than moon that despite of moon being closer to earth force due to sun is greater

But in case of Tidal Pull it is observed that it is inversely proportional to cube of distance i.e.

Where FT is tidal force or Tidal pull and r is the distance

So here as the distance increases tidal pull decreases by a huge amount and hence since sun is very far away as compared to moon tidal effect of moon’s Pull is greater.

A. Acceleration due to gravity decreases with increasing altitude, as it varies inversely to the square of distance from centre of earth and is given by relation

Where, g is the acceleration due to gravity

G is universal gravitational Constant

M_e is mass of Earth

r is distance of the point from center of earth (point must be on or above surface of earth not inside)

As can be seen in the figure

so as the distance from center of earth r or Altitude increases, the acceleration due to gravity decreases

B. Acceleration due to gravity decreases with increasing depth, as though distance of center of earth from the point is decreasing but mass of earth is also decreasing as less section of earth’s mass will contribute to Gravity as can be seen in the figure

If at a depth d inside surface of earth, the acceleration due to gravity is given as

g’ = g(1-d/R)

where, g’ is acceleration due to gravity at a depth d inside surface of earth, R is the Radius of earth and g is acceleration due to gravity on surface of earth

so as we can see as the depth inside surface d increases, value of d/R increases and (1-d/R) decreases and becomes less than 1, and hence we get

g’ < g inside surface of earth

C. Acceleration due to gravity is independent of mass of body as it is given by the relation

Where, g is the acceleration due to gravity

G is universal gravitational Constant

Me is mass of Earth

R is the radius of earth

as we can see it does not include any term of mass of body, so acceleration of gravity has same value for all bodies and is independent of mass of other body

D. The formula –G Mm(1/r₂ – 1/r₁) is more accurate than the formula mg(r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth, as acceleration due to gravity varies with distance from centre from earth

Potential energy assuming acceleration due to gravity to be constant is given as

V = mgr

Where V is the gravitational Potential Energy of body of mass m at a distance r from centre of earth, g is acceleration due to gravity

So at distance r₁ from centre of earth gravitational potential energy will be

V₁ = mgr₁

at distance r₂ from centre of earth gravitational potential energy will be

V₂ = mgr₂

So difference in potential energy is

V₂ – V₁ = mgr₂ – mgr₁ = mg(r₂ - r₁)

But as we know Acceleration due to gravity decreases with increasing altitude, as it varies inversely to the square of distance from centre of earth and is given by relation

Where, g is the acceleration due to gravity

G is universal gravitational Constant

M_e is mass of Earth

r is distance of the point from center of earth

so g has different value at both the points i.e. the difference in potential energy is not accurate

The accurate relation for Gravitational Potential energy of a body of mass m at any point above surface of earth is given by relation

V = -GMm/r

Where V is the gravitational Potential Energy of body of mass m at a distance r from centre of earth and M is the mass of earth,

So at distance r₁ from centre of earth gravitational potential energy will be

V₁ = -GMm/r₁

at distance r₂ from centre of earth gravitational potential energy will be

V₂ = -GMm/r₂

So difference in potential energy is

V₂ – V₁ = -GMm/r₂ – (-GMm/r₁)

= –G Mm(1/r₂ – 1/r₁)

This is more accurate formula for change in gravitational potential energy

Now planet revolves in orbit around sun due to attractive gravitational force between planet and sun, from Kepler’s third law of planetary motion we know that square of time period one complete rotation of planet T around the sun is proportional to the cube of mean distance (average distance) between the planet and sun R

T²∝ R³

Le the Time taken by the earth for one complete revolution be T_e

T_e = 1 Year

Let mean distance of the earth from the sun or orbital radius be R_e

Since the planet is moving twice as fast as earth the time taken by the planet to complete one complete revolution is half of that taken by earth

Let the Time period of revolution of the planet be T_p, then we have

T_P = 1/2T_e = 1/2 year

Let, the orbital radius of this planet be R_P

Now, according to the Kepler’s third law of planetary motion, we have

And

Using both equations we get the relation

Or we can say

Simplifying we get the relation for the orbital radius of the planet as

Since planet took half as much time as earth so we have

T_p/T_e = 1/2

i.e. the radius of the planet is

Therefore, the radius of the orbit of this planet is 0.63 times the radius of the orbit of the Earth.

If a body orbits around other heavier body due to gravitational force of attraction then we have a relation between mass of heavier body, time period of revolution and, radius of the orbit as

Where M is the mass of heavier body, R is the radius of orbit, G is universal gravitational Constant and T is time period of revolution

Here we will consider two cases, motion of Satellite Io around Jupiter and motion of earth around sun

So for motion of Io around Jupiter we have

Where, M_j is the Mass of Jupiter

G is Universal gravitational constant

R_Io is radius of Io’s orbit, we are given

R_Io = 4.22 × 10⁸ m

T_Io is the time period of revolution of Io around Jupiter

TIo = 1.769 days

So for motion of Earth around Sun we have

Where, Ms is the Mass of sun

G is Universal gravitational constant

R_e is radius of Earth’s orbit

R_e = 1 AU = 1.496 × 10¹¹ m

T_e is the time period of revolution of Earth around sun

T_e = 365.25 days

Diving equation of mass of sun with equation of mass of Jupiter to compare

So putting values of R_e, R_Io, T_Io, T_e in above equation we get

NOTE: Time periods are not converted to SI units second, but are in years, but would not make any difference to result because we are taking ration and units and converting factors are going to be cancelled out ultimately

Solving above equation we get

Or we can say

M_s ≈ 1000M_j

i.e. mass of Sun is nearly 1000 times mass of Jupiter

There are huge Stars in galaxy which together constitute a very huge mass and centre point of all concentrated mass galactic centre, we assume Star revolve around galactic centre in the same way in which planets revolve around sun we can assume all starts in milky way to constitute one heavenly body assuming all the mass to be concentrated at a single point Galactic centre and other heavenly bodies like stars revolve around galactic centre so we can treat distance of star from milky way as the orbital radius of revolving star

We know If a body orbits around other heavier body due to gravitational force of attraction then we have a relation between mass of heavier body, time period of revolution and, radius of the orbit as

Where M is the mass of heavier body, R is the radius of orbit, G is universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

and T is time period of revolution

So rearranging above equation we get time period of revolution as

We are give total mass of Mass of Galaxy as

M = 2.5 × 10¹¹ solar mass

Now 1 solar mass is mass of sun which is

M_s = 2 × 10³⁰ Kg

So total mass of the galaxy is

M = 2.5 × 10¹¹ × M_s

= 2.5 × 10¹¹ × 2 × 10³⁰ Kg

= 5 × 10⁴¹ Kg

we are given distance of star from galactic center which is Radius of Milky way or orbital radius for star that is

R = 50000 ly = 5 × 10⁴ ly

We know that one light year is distance travelled by light in one year

And distance is given as

S = V × t

Where S is the distance covered in time t moving at a speed V

Speed of light is

V = 3 × 10⁸ m/s

Converting time of 1 year into seconds

1 year = 365.25 days

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

So we get time of 1 year as

t = 365.25 × 24 × 60 × 60 s

so we get distance of 1 light year as

S = 3 × 10⁸ m/s × 365.25 × 24 × 60 × 60 s

i.e 1 light year = 9.46 × 10¹⁵ m

Therefore, the radius of milky way is

R = 5 × 10⁴ × 9.46 × 10¹⁵ = 4.73 x 10²⁰ m

So putting values of R, M, G in equation to find time period

= 4.246 × 10¹⁵ s

Converting into years

We already know

1 year = 365.25 × 24 × 60 × 60 s

So 1 s = 1/(365.25 × 24 × 60 × 60) Year

So we get time period in the year as

So star will take 1.34 × 10⁸ Years to complete one revolution

A. If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its Kinetic Energy

As total energy of a body is sum of kinetic energy and potential energy, kinetic energy of a body is always positive and depends upon speed of body, potential energy is negative and decreases as we move away from orbiting planet and at infinite distance from planet it becomes 0, the total energy of bound system is negative i.e. if satellite is at finite distance from earth and is under influence of gravitational force of earth its Total energy will be negative and is half of the potential energy, we follow the relation

T = V/2 = -K

Where T is the total energy, V is potential energy and K is kinetic energy

Potential energy and Total energy are negative, the magnitude of total energy is equal to the magnitude of kinetic energy but the sign is opposite as kinetic energy is positive.

B. The energy required to launch an orbiting satellite out of earth’s gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence

As body will be out of earth’s influence when its total energy is Zero or positive and we know the Total energy of a body is the sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy kinetic energy of a body which is always positive and depends upon the speed of the body, potential energy is negative and decreases as we move away from orbiting the planet and at an infinite distance from the planet

Kinetic Energy is given as

Where K is the kinetic energy of a body of mass m, moving with velocity v

The potential energy of a body above the surface of the earth is given as

U = -GMm/r

Where U is the gravitational Potential Energy of body of mass m at a distance r from the centre of the earth and M is the mass of earth G is universal gravitational Constant

If Body have to move out of Earth’s influence its total energy should be positive i.e.

T ≥ 0

Or

so we need to give it energy which will increase its speed and increase its kinetic energy and total energy will be positive , now Both bodies are at same height from earth surface so have same value of negative potential energy Stationary body has zero kinetic energy since its velocity is zero but a body already orbiting earth has some speed i.e. some kinetic energy i.e. its total energy is already less negative than stationary object so consequently less energy is needed to be provided to make its total energy zero consequently i.e. less energy is needed to energy required to launch an orbiting satellite out of earth’s gravitational influence as compared to stationary object at same height

Escape speed is the minimum speed a body must acquire in order to move out of earth’s gravitational field or escape from earth’s influence. This happens when total energy body becomes zero or positive and is in the unbound state.

(a)No, Escape speed does not depend upon the mass of the body as escape speed is given by the relation

Here, VS is escape velocity of a body at a distance of R from the centre of earth G is universal Gravitational Constant, Me is the mass of earth

So as we can see the relation of escape velocity is independent of the mass of body itself.

(b) No, Escape speed does not depend upon location from where it is projected assuming the condition that earth is a perfect sphere and radius of earth’s surface is almost same everywhere, because then according to relation of escape velocity, the distance from centre of earth R will be a constant value irrespective of latitude and longitude hence the escape speed is invariant of location.

(c) No, Escape speed does not depend upon the direction of projection as we take any direction value of gravitational Constant G, the mass of earth Me and Radius at the surface of earth R will be constant so escape speed will be constant. Though if we take earth’s spin into account as well then the value will depend on direction though we are neglecting it here.

(d) Yes, Escape speed depends on the height of the location from where the body is launched as we can see in the relation escape velocity varies inversely to the square root of the distance from the centre of the earth

Where Vs is escaped speed and R is the distance from the centre of the earth

Now as the height of body from the surface of the earth increases its distance from the centre of earth increases and Escape velocity decreases.

Suppose a comet orbits around sun in elliptical orbit , so its motion will be in accordance with Kepler's laws , now suppose at any instant planet was at a point A at a distance of r from sun and after a while it moved to a point B at a distance R’ from sun in this motion it subtended an angle 𝜽 with the sun and two position and its speed changed from v to v’ direction of motion is always tangential to orbit

The situation has been shown in figure

(a) Linear speed of the comet is not constant, and its speed increases when its closer to sun and decreases when its farther from sun and since, according to Kepler's second law comet must sweep equal areas in equal interval of time, i.e. area made between sun and line joining two positions of comet must be same in the same interval of time, but sun is not equidistant from planet so speed of planet changes so that area swept remains same.

(b) Now angular speed is also not constant as it is the rate of change of angle which is subtended between the two positions of comet and sun, since planet, speed changes so it covers unequal angles in equal interval of time so its angular speed varies.

(c) Angular momentum is conserved or same because there is no external torque acting on the comet as a force acting due to sun on a comet is parallel to line joining sun and planet or it is a central force so torque on a comet is zero and when no external torque acts on a body its angular momentum is conserved.

(d) Kinetic energy is not same or conserved as the speed of comet is varying and kinetic energy depends only on the speed of particle We know Kinetic Energy is given as

Where K is the kinetic energy of a body of mass moving with speed v, so as speed changes kinetic energy also changes

(e) Potential energy is also not conserved or same as it depends upon separation of two bodies and as the orbit is elliptical the distance between comet and sun changes continuously so potential energy is also not conserved as potential energy is given as

U = -Gm1m2/R

Where U is the potential Energy of a body of mass m1 at a point at a distance R from a center of mass of Body of mass m2, G is universal Gravitational constant

Since distance R of the comet of mass m1 changes from the sun of mass m2 changes continuously potential energy also changes.

(f) The total energy of the system is the sum of kinetic energy and potential energy and is conserved always according to the law of conservation of energy, so even kinetic energy and potential energy varies, the total energy remains same. i.e. both are interconverted to each other.

(b) swollen face, (c) headache, (d) orientational problem are likely to afflict an astronaut in space

Swollen feet is a situation when due to excess body mass , resulting in excessive body weight applies force on feet which results in causing fluid to build up in feet, legs and ankles , but in space this problem will not afflict an astronaut in space because , there is no gravity so there will be no force on feet due to body weight as body weight will be zero though mass will be same we know weight is given as

W = mg

Where W is the weight of a body of mass m and g is acceleration due to gravity

In space there is no gravity so

g = 0

i.e. body weight is zero so there is no effect of swollen feet.

as there is no air in space and hence no air pressure, blood circulation with an increase in body and face as there is no external pressure to counter it, so it may cause the face to swell and also lead to a headache as blood circulation in brain too increases

Orientation is a function of mind involving awareness place time person, space also has a definite orientation (definite position of one point with respect to another) so if a person has orientation problem i.e. brain is not working properly regarding positions and direction this will definitely afflict the person in space

(iii) c

The gravitational intensity due to a body at a point is the force acting per unit mass on a body at that point or we can say it is force on a unit mass, it’s a vector quantity and its direction is same as direction of force on unit mass, so in absence if any external gravity if we put a point mass at centre of hemispherical shell it will be pulled downwards so Direction of field intensity is Downwards

Further to prove it if we take a uniform spherical shell of the same radius, at each point inside the hollow spherical shell, Gravitational field intensity will be zero as Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.

Now the shell is spherically symmetric so if the upper half of shell, pulls a point mass towards it with the magnitude of Force F so same with the amount of force lower half of shell will pull it as shown in figure

So if we remove the upper half of sphere there will force acting on point mass will be in a downward direction only, so that will be the direction of field intensity as shown by (c) in the figure Below

(ii) e

The gravitational intensity due to a body at a point is the force acting per unit mass on a body at that point or we can say it is force on a unit mass, it’s a vector quantity and its direction is same as direction of force on unit mass, so in absence if any external gravity if we put a point mass at any point inside hemispherical shell it will be pulled downwards so Direction of field intensity is downwards

Further to prove it if we take a uniform spherical shell of the same radius, at each point inside the hollow spherical shell, Gravitational field intensity will be zero as Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.

Now the shell is spherically symmetric so if the upper half of shell, pulls a point mass towards it with the magnitude of Force F so same with the amount of force lower half of shell will pull it as shown in figure

So if we remove the upper half of sphere there will force acting on point mass will be in a downward direction only, so that will be the direction of field intensity as shown by (e) in the figure Below

The net gravitational force on the rocket will be zero when attractive gravitational force on Rocket due to sun is equal in magnitude to attractive gravitational force by Earth, so it is between earth and sun, suppose distance between earth and sun or the orbital radius is R, and rocket is at a distance r from earth so it is at (R-r) distance from sun and attractive pull balances attractive pull of earth

The situation has been depicted in the figure

The gravitational force on Rocket due to sun Fs is equal in magnitude to attractive gravitational force by Earth Fe and opposite in direction as can be seen in the figure

we know gravitational force on a body is given as

Where F is the gravitational force

G is universal gravitational Constant

m₁ is mass of the first body

m₂ is the mass of the second body

and r is the distance between the two bodies

Now gravitation force on rocket due to earth will be

Where Me and Mr are masses of earth and rocket and r is a separation between them

Similarly, gravitation force on rocket due to Sun will be

Where Ms and Mr are masses of Sun and rocket and separation between Sun and rocket is (R-r)

Since both forces should be equal in magnitude, equating both

F_e = F_s

i.e.

solving and canceling terms we get

Or we can say

We are given

Mass of the sun, M_s = 2×10³⁰ kg

Mass of the earth, M_e = 6×10²⁴ kg

The distance between Earth and Sun or orbital Radius

R = 1.5 × 10¹¹ m

So putting these values to find the distance between earth and rocket r

577.3r = 1.5 × 10¹¹ m – r

578.3r = 1.5 × 10¹¹ m

r = (1.5 × 10¹¹ m)/ 578.3 = 2.59 × 10³ m

so the rocket is at a distance of 2.59 × 10³ m from earth’s centre

We know that earth revolves around sun in elliptical orbit due to gravitational force applied by sun, we also know that in one complete revolution earth takes 365.25 days and we are given mean orbital radius of earth around sun, now we know that If a body orbits around other heavier body due to gravitational force of attraction then we have a relation between mass of heavier body, time period of revolution and, radius of the orbit as

Where M is the mass of the heavier body, R is the radius of the orbit, G is universal gravitational Constant and T is time period of revolution

Now here unknown mass is mass of the sun, M =?

Orbital radius of earth around sun, R = 1.5 × 10⁸ km

Time period of one revolution, T = 365.25 days

Converting into seconds

1 day = 24 hours

1 hour = 60 minutes

1 minute = 60 seconds

So we get time of 1 year as

T = 365.25 × 24 × 60 × 60 s = 3.15 × 10⁷ s

Value of universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

Putting these values in above relation

Solving we get the mass of sun as 2.0 × 10³⁰ Kg

Earth and Saturn both revolve in definite orbits around sun due to gravitational force of attraction due to sun now when a planet revolves in orbit around sun due to attractive gravitational force between planet and sun, from Kepler’s third law of planetary motion we know that square of time period one complete rotation of planet T around sun is proportional to cube of mean distance (average distance) between planet and sun R

T²∝ R³

Le the Time taken by the earth for one complete revolution be T_e

T_e = 1 Year

Let mean distance of the earth from the sun or orbital radius be Re

Time is taken by the Saturn for one complete revolution or its time period of revolution around the sun be T_s

We are given that a Saturn’s year is 29.5 times the earth year i.e. time period of Saturn is 29.5 times that of earth

i.e. T_s = 29.5 T_e

Let, the orbital radius of this Saturn be R_s

Now, according to the Kepler’s third law of planetary motion, we have

And

Using both equations we get the relation

;

Or we can say

Simplifying we get the relation for the orbital radius of Saturn as

now we have

T_s = 29.5 Te

i.e. T_s/T_e = 29.5

i.e. radius of planet is

Putting the value of the orbital radius of the earth

R_e = 1.50 × 10⁸ km

we get the distance of Saturn from the sun as

R_p = 9.54 × 1.50 × 10⁸ km = 1.43 × 10⁹ Km = 1.43 × 10¹² m

So orbital radius of Saturn is 1.43 × 10⁹ Km or we can say it is at a distance of 1.43 × 10¹² m from sun

The weight of a body is the force acting one it due to earth’s gravity and is the product of mass and acceleration due to gravity is given as

W = mg

Where W is the weight of a body of mass m on the surface of the earth, g is acceleration due to gravity on the surface of the earth

Now as the height from the surface of the earth increases acceleration due to gravity decreases, so the weight of the body also decreases i.e.

W_h = mg_h

Where W_h is the weight of a body of mass m kept on h height above the surface of the earth, Let g_h be the acceleration due to gravity at a height h above the surface of the earth

Now acceleration due to gravity above surface h varies according to the relation

Here,

g_h is the acceleration due to gravity at a height h above the surface of the earth, g is acceleration due to gravity on earth’s surface and R is Radius of the Earth

The situation has been shown in figure

In this case, height is equal to half of the radius of earth i.e.

H = R/2

Therefore,

On solving further we get

i.e. we get

g_h = 4/9 g

So acceleration due to gravity at a height R/2 above the surface of the earth is 4/9 times to that on the surface of the earth

So weight at that height will be

W_h = m(4/9g) = 4/9 mg

i.e.

W_h = 4/9 W

So weight will be 4/9 times weight at the surface of the earth

We are given weight on the surface of the earth as

W = 63 N, so we have

W_h = 4/9 × 63N = 28 N

So the weight of the body or gravitational force on it due to the earth at a height equal to half the radius of the earth will be 28 N

The weight of a body is the force acting one it due to earth’s gravity and is the product of mass and acceleration due to gravity is given as

W = mg

Where W is the weight of a body of mass m on the surface of the earth, g is acceleration due to gravity on the surface of the earth

Now as the body goes inside surface of earth towards the centre of earth acceleration due to gravity decreases, so the weight of the body also decreases i.e.

W_d = mg_d

Where W_d is the weight of a body of mass m kept at a depth d inside the surface of the earth, Let g_d be the acceleration due to gravity at a at a depth d inside the surface of the earth

Now acceleration due to gravity at a depth d inside the surface of the earth is given by the relation

Where g_d is the acceleration due to gravity at a depth d inside the surface of the earth, g is acceleration due to gravity on earth’s surface and R is Radius of the Earth

The situation has been shown in figure

Now Let g_d be the acceleration due to gravity half way down to the centre of the earth and g be the acceleration due to gravity on the surface of the earth.

If R is the radius of the Earth and mass in half way down the earth then we have the depth of body from the surface of earth

d = R/2

Then using

We get

g_d = g/2

i.e. acceleration due to gravity half way down the surface of the earth is half of that on the surface of the earth

so the weight of the body at this depth will be

W_d = mg_d

W_d = m(g/2) = 1/2 mg

i.e. W_d = 1/2 W

now we are given the weight of the body on the surface of the earth as

W = 250 N

i.e. we have

W_d = 250/2 N

W_d = 125 N

So, the weight of the body at a depth equal to half the radius of the earth is 125 N.

Rocket fired from surface of Earth and will have kinetic energy(Positive) by virtue of speed which is taking it away from surface of Earth, and gravitational potential energy(Negative) is trying to pull it towards the centre of Earth, total energy of the rocket is the sum of kinetic energy and gravitational potential energy , and total energy is conserved , i.e. will remain same

Now as rocket is fired upward Earth’s gravity will apply force on it pull it back, and speed of rocket will decrease and kinetic energy also, potential energy on another hand will increase and become less negative and at a maximum height h above the surface of Mars at a distance R from centre of mars of radius r, all kinetic energy will be converted to potential energy, and rocket will come to rest and then will revert back towards surface of Mars

The situation has been shown in figure

We know Kinetic Energy is given as

Where K is the kinetic energy of a body of mass moving with speed v, here we are given

The initial speed of Rocket as

v = 5 km s^–1 = 5000 ms^-1

let Mass of the rocket be m

so kinetic energy is

Now Gravitational potential energy is negative and decreases as we move away from System and at an infinite distance from the planet

Now Gravitational potential at a point is given by the relation

U = -GMm/R

Where U is the potential Energy of a body of mass m at a point at a distance R from a center of mass of Body of mass M, G is universal Gravitational constant

Initially, Rocket is at the surface of Earth so its distance from the centre of mars is equal to the radius of Earth i.e.

R = r = 6400 km = 6.4 × 10⁶ m

let Mass of the rocket be m

mass of planet Earth is

M = 6×10²⁴ kg

Value of universal gravitational constant

G = 6.67 × 10^-11 Nm²Kg^-2

So putting these value we get initial potential energy U_i as

= -6.25m × 107 J

Now finally after reaching a height h, the rocket will come to rest and its velocity will be 0, i.e.

v = 0 m/s

so final kinetic energy will also be zero i.e.

K_f = 0 J

Now after reaching a height h, the distance of rocket from the centre of mars will be

R = r + h

So final potential energy U_f will be

i.e

The total energy of a body is the sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy

So total initial energy will be

T_i = U_i + K_i

i.e. T_i = -6.25m × 10⁷ J + 25m/2 × 10⁶ J

= - 50.0m × 10⁶ J

So Total Final Energy will be

T_f = U_f + K_f

i.e.

Now by the law of conservation of energy, we know the total initial energy of rocket must be equal to the total final energy of the rocket i.e.

T_i = T_f

i.e.

now cancelling out a negative sign and term m from both sides and solving we get

After cross multiplication we get

320 × 10⁶ + 50h = 4 × 10⁸

i.e. 50h = 400 × 10⁶ – 320 × 10⁶ = 80 × 10⁶

so we get h as

h = 80 × 10⁶/50 = 1.6 × 10⁶ m

So the rocket will reach a height of 1.6 × 10⁶ m above the surface of Earth

Its Distance from centre of earth will be

R = r + h

Where r is the radius of earth which is

r = 6.4 × 10⁶ m

so we get

R = 6.4 × 10⁶ m + 1.6 × 10⁶ m

= 8 × 10⁶ m

So satellite is at a distance of 8 × 10⁶ m from surface of earth

Escape speed is that after attaining which a body moves just out of the earth’s Gravitational influence

As body will be out of earth’s influence when its total energy is Zero or positive and we know the Total energy of a body is the sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy kinetic energy of a body which is always positive and depends upon the speed of the body, potential energy is negative and decreases as we move away from orbiting the planet and at an infinite distance from the planet

Kinetic Energy is given as

Where K is the kinetic energy of a body of mass m, moving with velocity v

The potential energy of a body above the surface of the earth is given as

U = -GMm/R

Where U is the gravitational Potential Energy of body of mass m at a distance R from the centre of the earth and M is the mass of earth G is universal gravitational Constant

If Body have to move out of Earth’s influence its total energy should be positive i.e.

T ≥ 0

Or

Let m be the mass of the projectile and V_e be the escape speed from surface of earth. So kinetic energy of a body at escape speed K_e is

K_e = 1/2 m V_e²

when distance of body from centre of earth will be equal to radius of earth, Potential energy of particle at surface of earth is

U = -GMm/r

i.e. for a body to just escape total energy should be zero i.e.

i.e. K_e = 1/2 m V_e² = -(-GMm/r)

or U = -K_e

so potential energy of a body at surface of earth is equal to negative of kinetic energy at escape velocity

We are given initial speed V_i is three times escape speed i.e.

V_i = 3Ve

Let m be the mass of the projectile, then its initial kinetic energy will be

K_i = 1/2 mV_i² = 1/2 m(3V_e)²

= 9(1/2 m V_e²) = 9 K_e

So initial kinetic energy of projectile is 9 times kinetic energy at escape speed

Now initial potential energy of projectile when it is at surface of earth will be

U_i = -GMm/R

We know it will be equal to negative of the kinetic energy of the same particle at escape velocity

U_i = -K_e

So total initial energy of projectile will be

T_i = K_i + U_i

i.e. T_i = 9 K_e + (-K_e) = 8 K_e

so total initial energy of projectile will be 8 times its kinetic energy at escape velocity

now finally when projectile will be at an infinite distance, its potential energy will be zero

U_f = 0

And let us assume particle of mass m has gained a velocity V_f, so the final kinetic energy of projectile will be

K_f = 1/2 mV_f²

So total final energy will be

T_f = K_f + U_f

i.e. T_f = 1/2 mV_f²

using the law of conservation of energy we know total initial energy must be equal to total final energy so we have

T_i = T_f

Or we can say

8(K_e) = 1/2 mV_f²

8 x (1/2 mV_e²) = mV_f²

On solving we get the relation between final speed V_f and Escape speed velocity V_i as

V_f² = 8V_e²

Or

Now we know escape velocity is

V_e = 11.2 km/s = 11.2 × 10³ m/s

the final velocity of projectile Vf far away from earth will be

i.e. V_f = 31.68 × 10³ m/s = 31.68 km/s

so final velocity of Projectile at distance far away from earth is 31.68 km/s

Suppose a satellite is orbiting earth’s surface, Seattleite will be out of earth’s influence when its total energy is Zero or positive and we know Total energy of a body is sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy kinetic energy of a body and is always positive, and depends upon the speed of the body, potential energy is negative and decreases as we move away from orbiting the planet and at an infinite distance from the planet

Kinetic Energy is given as

Where K is the kinetic energy of a body of mass m, moving with velocity v

The potential energy of a body above the surface of the earth is given as

U = -GMm/R

Where U is the gravitational Potential Energy of body of mass m at a distance R from the centre of the earth and M is the mass of earth G is universal gravitational Constant

If Body have to move out of Earth’s influence its total energy should be positive i.e.

T ≥ 0

Or

so we need to give it energy which will increase its speed and increase it's kinetic energy and total energy will be positive assume earth to be a sphere of radius r, and satellite is orbiting the earth at a distance h above the surface of the earth at a distance R from the centre of the earth as Shown in figure

So we have

R = r + h

So Total energy T of the satellite orbiting at height h above the surface of the earth will be

T = U_i + K_i

Where U_i and K_i are initial kinetic and potential energy of the satellite respectively i.e.

where M is mass of earth, m is mass of satellite, r is Radius of earth, and v_o is the orbital velocity of the satellite

The total energy of satellite will be negative and is half of the potential energy, we follow the relation

T = U/2 = -K

Where T is the total energy, U is potential energy and K is kinetic energy

So we can see kinetic energy is (-1/2) times potential energy

So we get

K_i = -U/2 = GMm/(2(r+h)

i.e.

Suppose E amount of energy is expended on a rocket to make its total energy 0

i.e.

E + T = 0

Or E = -T

So Energy expended to rocket the satellite out of the earth’s gravitational field will be negative of initial total energy

Where M is mass of earth which is

M = 6.0 × 10²⁴ kg

m is mass of satellite which is

m = 200 kg

G is universal gravitational constant

G = 6.67 × 10^–11 N m2 kg^–2

r is the radius of the earth

r = 6.4 × 10⁶ m

h is the initial height of satellite above the surface of the earth

h = 400 Km = 4 × 10⁵ m

so putting values in above equation, we get

So The energy expended to move the satellite out of earth’s gravitational field is 5.9 x 10⁹ J

Now let us assume both stars to be initially at rest and come towards each other due to the gravitational force of attraction and as they come closer their speed keeps on increasing, suppose they are initially at a distance of d

d = 10⁹ km = 10¹² m

For head-on collision, the least distance between the stars will be equal to twice of their radius

The Initial situation and when they are just going to collide has been displayed in the figure below

For head-on collision, the least distance between the stars will be equal to twice of their radius, we are given the radius of each star as

r = 10⁴ km = 10⁷ m

so least distance or final between the stars will be

d’ = 2r = 2 × 10⁷ m

we will use the law of conservation of energy

The total energy of a body is the sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy

We know Kinetic Energy is given as

Where K is the kinetic energy of a body of mass moving with speed v and potential energy is given as

U = -Gm₁m₂/R

Where U is the potential Energy of a body of mass m₁ at a point at a distance R from a center of mass of Body of mass m₂, G is universal Gravitational constant

Initially, Both the Stars are assumed to be at rest i.e their speed is

v = 0

so initial Kinetic energy of both the stars will be zero so the total kinetic energy of the system consisting two stars will also be zero

i.e. K_i = 0

Initially, potential energy of the system will have a maximum value when the separation between stars will be maximum

now masses of both the stars are equal i.e.

m¹ = m² = m = 2 × 10³⁰ kg

value of the universal gravitational constant is

G = 6.67 × 10^-11 Nm²Kg^-2

And the initial separation between the stars is

R = d = 10¹² m

so the initial gravitational potential energy of the system U_i will be

= -26.68 × 10³⁷ J

So total initial energy of the system will be

T_i = U_i + K_i

i.e. T_i = -26.68 × 10³⁷ J + 0 J

= -26.68 × 10³⁷ J

Now finally let us assume both stars gained speed v moving towards each other and are about to collide, so the final kinetic energy of the system will be equal to the sum of kinetic energy of both the stars i.e.

K_f = 1/2 mv² + 1/2 mv² = mv²

Where m is the mass of each star, we know

m = 2 × 10³⁰ kg

so the final kinetic energy of the system will be

K_f = 2 × 10³⁰ × v²

Finally when stars will come close and separation between them will be minimum potential energy will also be at its minimum value

We know final separation between the stars is

R = d’ = 2 × 10⁷ m

now masses of both the stars are equal i.e.

m₁ = m₂ = m = 2 × 10³⁰ kg

value of the universal gravitational constant is

G = 6.67 × 10^-11 Nm²Kg^-2

so Final gravitational potential energy of the system U_f will be

= -13.34 × 10⁴² J

So total final energy of the system will be

T_f = U_f + K_f

i.e. T_f = -13.34 × 10⁴² J + 2 × 10³⁰ × v²

NOTE:while calculating total kinetic energy of the system we added individual kinetic energies of both the stars but we did not do this for potential energy as potential energy of a body due to another body is the potential energy of the system and taking any body as reference (P.E. of star 1 w.r.t star 2 or P.E. of star 2 w.r.t. star 1) when we calculate potential energy value is same

now as we can observe potential energy of the system has been lost and is converted to kinetic energy but according to the law of conservation of energy total energy will be same i.e.

T_i = T_f

So equating values

-26.68 × 10³⁷ J = -13.34 × 10⁴² J + 2 × 10³⁰ Kg × v²

Solving further

2 × 10³⁰ Kg × v² = 13.34 × 10⁴² J – 0.00026 × 10⁴² J

v² = 6.66 × 10¹² m²s^-2

i.e

So, we get final speed of both the stars as 2.58 × 10⁶ ms^-1

Let us consider two spheres each of mass M and radius r are placed at a distance d apart from each other on a horizontal table,

Here a mass of both the spheres is

M = 100 Kg

Here Radius of both the spheres is

r = 0.10 m

The separation between both the spheres is

d = 1.0 m

suppose sphere 1 is on left side of table and Sphere 2 is on right side of table, suppose a body is placed at the midpoint of line joining two spheres, Since gravitational force is always attractive in nature so let Gravitational Force due to sphere 1 be F₁ and it will be in left direction, and force due to sphere 2 is F₂ and will act on right side

The situation has been shown in the figure

We know gravitational force on a body is given as

Where F is the gravitational force

G is universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

M is mass of the first body

m is the mass of the second body

and R is the distance between the two bodies

now here if Body is kept at the midpoint of the line joining two spheres then the separation between a body at the midpoint and the two spheres will be same and equal to half the distance between two spheres suppose R₁ is a distance of the first sphere from midpoint and R₂ is a distance of the second sphere from midpoint then we have

R₁ = R₂ = R = d/2 = 1/2 = 0.5m

now we get that magnitude of force on a body kept at midpoint by sphere 1 and sphere 2 is same as for both the forces F₁ and F₂, mass of first body(Sphere) M, mass of the second body(body at midpoint) m, separation between both the bodies R is same and Universal gravitational constant is same , i.e. magnitude of forces

F₁ = F₂ = GMm/R²

Now magnitude of both the forces will be exactly same but directions will be opposite, and force is a vector quantity, so forces will be added vectorially and we know Force of equal magnitude and opposite direction result in Zero as both cancel out each other, Magnitude of the net force is

F = F₁ – F₂ = 0 N

So net force at the centre is 0 N

Now Gravitational potential at a point is given by the relation

V = -GM/R

Where V is the potential of a point at a distance R from a Body of mass M, G is universal Gravitational constant

Potential is a scalar quantity, so it does not have a direction and is added directly not net potential at the centre will be

V = V₁ + V₂

Where V₁ is the potential due to the first body and V₂ is the potential due to the second body

We know potential due to the first sphere will be given as

V₁ = -GM₁/R₁

Where M1 is the Mass of the first sphere

M₁ = M = 100 Kg

R₁ is a distance of the first sphere from Midpoint

R₁ = R = 0.5 m

G is universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

Now putting the values in above equation, we get

similarly, potential due to the second sphere will be given as

V₂ = -GM₂/R₂

Where M₂ is the Mass of the Second sphere

M₂ = M = 100 Kg

R₂ is a distance of the Second sphere from Midpoint

R₂ = R = 0.5 m

G is universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

Now putting value we get

So net potential at mid-point is

V = V₁ + V₂

= -1.334 × 10^-8 Jkg^-1 + (-1.334 × 10^-8 Jkg^-1)

= -2.668 × 10^-8 Jkg^-1

If a body is placed at Midpoint it will be in equilibrium as net force acting on it will be Zero , but it be in unstable equilibrium as if it is displaced slightly to left or right it will be pulled towards the sphere which is at less distance from it and not come back to its mean position this will happen because gravitational force of attraction is inversely proportional to distance so as distance between the two bodies decrease force of attraction increase and vice versa, so on even slight displacement force on body due to one sphere will increase and decrease due the other sphere which is at a greater distance and it will be pulled towards body at lesser distance and never come back to original position such an equilibrium condition is called unstable equilibrium.

Gravitational potential energy is the energy possessed by a body at some point due to its mass and its power to attract object due to its Gravity

We know Gravitational potential at a point is given by the relation

V = -GM/R

Where V is the potential of a point at a distance R from a Body of mass M, G is universal Gravitational constant

Now R is the Distance from the center of the body, so in case of a satellite at some height h from earth surface, the distance of earth’s satellite from the centre of earth R will be equal to sum of earth’s radius r and height from the surface of earth h

R = r + h

The situation has been shown in figure

We are given Radius of earth

r = 6400 Km

the height of satellite from Earth’s Surface

h = 36,000 km

so the net distance of point where the satellite is located to the center of the earth is

R = r + h = 6400 Km + 36,000 km

= 42400 Km = 4.24 × 10⁷ m

Mass of earth

M = 6.0 × 10²⁴ kg

Value of universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

Now putting the values in above equation we get potential at the point as

So the potential due to earth’s gravity at the site of this satellite is

-9.43 × 10⁶ Jkg^-1

Now if an object is on a star placed at equator it will move with the surface of planet and will actually cover a circular path with certain speed depending upon angular speed or rate of revolution of star, now due to movement in circular path with some speed the object will experience a outward centrifugal force pushing it out of the planet but gravitational force on the object due to star is pulling it inwards or towards the surface so for object to be stuck on the star the gravitational force must be greater than the centrifugal force

The situation has been explained in the figure

We know gravitational force on a body is given as

Where F is the gravitational force

G is universal gravitational Constant

G = 6.67 × 10^-11 Nm²Kg^-2

M is mass of the first body which in this case is a star

Mass of star is 2.5 times mass of the sun, and mass of the sun is

M_s = 2 × 10³⁰ Kg

So the mass of the star is

M = 2.5 × M_s = 2.5 × 2 × 10³⁰ Kg = 5 × 10³⁰ Kg

m is the mass of second body which in this case is the object, mass of an object is unknown let it be m

and R is the distance between the two bodies which here will be equal to the radius of star r as the object is on its surface i.e.

R = r = 12 Km = 1.2 × 10⁴ m

So putting values of M, m, R, and G we get the magnitude of the inward gravitational force on object F_G as

In inward direction, here m is mass of the object

Now body on the surface of the star is actually covering a circular path with a radius equal to the radius of the sun, the sun is revolving with a frequency of 1.2 revolutions per second so its rate of change of angle w.r.t center or angular frequency will be given as

ω = 2πf

where ω is the angular frequency of a body revolving with frequency f revolutions per second

here f = 1.2 rev/s

so we get the angular frequency of revolution as

ω = 2π × 1.2 rev/s = 7.539 rad/s

now centrifugal force Fc on the object which is undergoing circular motion in a circle of radius r with angular speed or frequency ω is given as

Fc = mrω²

Here radius of circular path or radius of the star is

r = 12 Km = 1.2 × 10⁴ m

The angular speed of the object is

ω = 7.539 rad/s

mass of the object is m, mass is unknown

so putting values we get outward centrifugal force as

Fc = m × 1.2 × 10⁴ m × (7.539 rad/s)² = m × 6.8 × 10⁵ N

In outward direction so clearly we can see comparing centrifugal force FC and Gravitational Force F_G that gravitational force is much greater as

F_G = m × 2.31 × 10¹² N, in an inward direction

F_C = m × 6.8 × 10⁵ N, in outward direction

Value of m i.e. mass of an object is same in both the equations

So we can say

F_G > F_C

i.e. effectively object is pushed inward towards the surface of the star so it will remain stuck on the surface of the star

Every Object possesses two kinds of energy potential energy due to its position and kinetic energy due to its speed, Whenever a body is bound to a system or another body like in this case Spaceship is bound to the solar system due to gravitational force applied by it in order to keep Spaceship inside the solar system , the total energy of such a object will be negative, whenever total energy of object will become positive or even zero it will escape from that system irrespective of all the forces being applied to it by the system.

The total energy of a body is the sum of kinetic energy and potential energy

T = K + V

Where T is the total energy, V is potential energy and K is kinetic energy kinetic energy of a body is always positive and depends upon the speed of the body, Gravitational potential energy is negative and decreases as we move away from System and at an infinite distance from the planet

Kinetic Energy is given as

Where K is the kinetic energy of a body of mass m, moving with velocity v

The potential energy of a body above the surface of the earth is given as

V = -GMm/r

Where V is the gravitational Potential Energy of body of mass m at a distance r from the centre of a body of mass M, G is universal gravitational Constant

Now here we are considering system to consist only of sun and Mars as all other bodies in solar system have very less mass compared to these, so total gravitation potential energy of body will sum of gravitational potential energy due to sun and due to Mars, now spaceship is on surface of Mars so its distance from surface of Mars will equal to radius of mars r, and its distance from sun will nearly be equal to distance of mars from sun i.e. radius of Mars orbit

As can be seen in the figure

So if we let the mass of spaceship be m and mass of mars be Mm then and radius of mars be r, then gravitational potential energy of object due to mars will be

V_m = -GMm_m/r

Similarly, mass of Sun be M_s then and radius of orbit of mars be R, then gravitational potential energy of object due to Sun will be

V_s = -GMs_m/R

So adding both potential energies we get total potential energy of the spaceship as

V = V_s + V_m

i.e. V = -GM_sm/R + (-GM_mm/r)

= -GM_sm/R -GM_mm/r = -Gm(M_s/R + M_m/r)

Spaceship is assumed to be at rest initially so its speed will be 0, i.e. v = 0 m/s

So its kinetic energy will be zero, so its total energy will be its total potential energy

so we need to give it energy which will increase its speed and increase its kinetic energy and total energy will be positive, if we supply E amount of energy so total energy of spaceship will be

T = E + V

Or we get

T = E + [-Gm(M_s/R + M_m/r)]

If Spaceship has to move out of System’s influence i.e. move out of the solar system its total energy should be positive i.e.

T ≥ 0

So taking the condition for spaceship to just escape the solar system

T = 0

i.e. E + [-Gm(M_s/R + M_m/r)] = 0

i.e. we get total energy that need to be supplied to the space ship is

E = Gm(M_s/R + M_m/r)

Now we know value of universal gravitational constant as

G = 6.67 × 10^-11 Nm²Kg^-2

Mass of the space ship is

m = 1000 kg

Mass of the sun is

M_s = 2 × 10³⁰ kg

mass of mars is

M_m = 6.4 × 10²³ kg

Radius of mars is

r = 3395 km = 3.395 × 10⁶ m

Radius of the orbit of mars is

R = 2.28 × 10⁸ km = 2.28 × 10¹¹ m

so putting the values of G, r, R, m, M_s, M_m in above equation we get

= 6.67 × 10^-11Nm²Kg^-2 × 1000Kg × (8.77 × 10¹⁸kgm^-1 + 1.88 × 10¹⁷ kgm^-1)

= 6.67 × 10^-8 Nm²Kg^-1 × 8.95 × 10¹⁸ kgm^-1

= 5.98 × 10¹¹ J

So, to launch the spaceship out of the solar system we need to supply 5.98 × 10¹¹ J of energy to Spaceship

Rocket fired from surface of mars will have kinetic energy(Positive) by virtue of speed which is taking it away from surface of mars but 20% of its initial energy is lost due to martian atmospheric resistance so its energy is less than initial energy provided to it , and gravitational potential energy(Negative) is trying to pull it towards the centre of mars , total energy of the rocket is the sum of kinetic energy remaining after atmospheric loss and gravitational potential energy , and total energy is conserved , i.e. will remain same

Now as rocket is fired upward Mars gravity will apply force on it pull it back, and speed of rocket will decrease and kinetic energy also, potential energy on other hand will increase and become less negative and at a maximum height h above the surface of mars at a distance R from centre of mars of radius r, all kinetic energy will be converted to potential energy, and rocket will come to rest and then will revert back towards surface of mars

The situation has been shown in figure

We know Kinetic Energy is given as

Where K is the kinetic energy of a body of mass moving with speed v, here we are given

Initial speed of Rocket as

v = 2 km s^–1 = 2000 ms^-1

let Mass of the rocket be m

so kinetic energy is

Now 20% of kinetic energy is lost so effective initial kinetic energy Ki is 80% of K, i.e.

= 1.6m × 10⁶ J

Now Gravitational potential energy is negative and decreases as we move away from System and at infinite distance from planet

Now Gravitational potential at a point is given by relation

U = -GMm/R

Where U is the potential Energy of a body of mass m at a point at a distance R from a centre of mass of Body of mass M, G is universal Gravitational constant

Initially Rocket is at the surface of Mars so its distance from centre of mars is equal to radius of mars i.e.

R = r = 3395 km = 3.395 × 10⁶ m

let Mass of the rocket be m

mass of planet mars is

M = 6.4 × 10²³ kg

Value of universal gravitational constant

G = 6.67 × 10^-11 Nm²Kg^-2

So putting these value we get initial potential energy U_i as

= -12.57m × 10⁶ J

Now finally after reaching a height h, the rocket will come to rest and its velocity will be 0, i.e.

v = 0 m/s

so final kinetic energy will also be zero i.e.

K_f = 0 J

Now after reaching a height h, the distance of rocket from centre of mars will be

R = r + h

So final potential energy U_f will be

i.e.

Total energy of a body is sum of kinetic energy and potential energy

T = K + U

Where T is the total energy, U is potential energy and K is kinetic energy

So total initial energy will be

T_i = U_i + K_i

i.e. T_i = -12.57 m × 10⁶ J + 1.6m × 10⁶ J

= -10.97 m × 10⁶ J

So Total Final Energy will be

T_f = U_f + K_f

i.e.

Now by law of conservation of energy we know total initial energy of rocket must be equal to total final energy of the rocket i.e.

T_i = T_f

i.e.

now cancelling out negative sign and term m from both sides and solving we get

After cross multiplication we get

37.34 × 10⁶ + 10.97h = 42.68 × 10⁶

i.e. 10.97h = 42.68 × 10⁶ - 37.34 × 10⁶ = 5.34 × 10⁶

so we get h as

h = 5.34 × 10⁶/10.97 = 0.495 × 10⁶ m

= 495 × 10³ m = 495 Km

So the rocket will reach a height of 495 Km above the surface of Mars.