**Gravitational Potential Energy**

- Potential energy is due to the virtue of position of the object.
- Gravitational Potential Energy is due to the potential energy of a body arising out of the force of gravity.
- Consider a particle which is at a point P above the surface of earth and when it falls on the surface of earth at position Q, the particle is changing its position because of force of gravity.
- The change in potential energy from position P to Q is same as the work done by the gravity.
- It depends on the height above the ground and mass of the body.

Stationary roller-coaster

__Expression for Gravitational Potential Energy__

**Case1**:- ‘g’ is constant.

- Consider an object of mass ‘m’ at point A on the surface of earth.
- Work done will be given as :
- W
_{BA}=F X displacement where F = gravitational force exerted towards the earth) - =mg(h
_{2}-h_{1}) (body is brought from position A to B) - =mgh
_{2}-mgh_{1} **W**_{AB}=V_{A}-V_{B}- where
- V
_{A}=potential energy at point A - V
_{B}= potential energy at point B

- V
- From above equation we can say that the work done in moving the particle is just the difference of potential energy between its final and initial positions.

**Case2**:-‘g’ is not constant.

- Calculate Work done in lifting a particle from r = r
_{1}to r = r_{2}(r_{2}> r_{1}) along a vertical path, - We will get , W=V (r
_{2}) – V (r_{1})

**Conclusion**: –

- In general the gravitational potential energy at a distance ‘r’ is given by :

**V(r) = -GM _{e}m/r + V_{o}**

- where
- V(r) = potential energy at distance ‘r’
- V
_{o}= At this point gravitational potential energy is zero. - Gravitational potential energy is ∝ to the mass of the particle.

**Gravitational Potential** ( Class 11 Physics Gravitation )

- Gravitational Potential is defined as the potential energy of a particle of unit mass at that point due to the gravitational force exerted byearth.
- Gravitational potential energy of a unit mass is known as gravitational potential.
- Mathematically:
**G**_{potential}= -GM/R

** Problem**: ( Gravitation )

Choose the correct alternative:

Acceleration due to gravity increases/decreases with increasing altitude.

Acceleration due to gravity increases/decreases with increasing depth. (assume the earth to be a sphere of uniform density).

Acceleration due to gravity is independent of mass of the earth/mass of the body.

The formula –G M_{m }(1/r_{2}– 1/r_{1}) is more/less accurate than the formula mg(r_{2}– r_{1}) for the difference of potential energy between two points r_{2}and r_{1} distance away from the centre of the earth.

__Answer:__

(a)Decreases

(b)Decreases

(c)Mass of the body

(d)More

Explanation:

- Acceleration due to gravity at depth h is given by the relation:

g_{h} = (1- 2h/R_{e})g

Where,

R_{e} = Radius of the Earth, g = acceleration due to gravity on the surface of the earth.

It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

- Acceleration due to gravity at depth d is given by the relation:

g_{d}=(1-d/R_{e})g

It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

- Acceleration due to gravity of body of mass m is given by the relation: g=GM/r
^{2}

Where,

G = Universal gravitational constant

M = Mass of the Earth

R = Radius of the Earth

Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

- Gravitational potential energy of two points r
_{2}and r_{1}distance away from the centre of the Earth is respectively given by:

V (r_{1}) = – G mM/r_{1}

V (r_{2}) = -G mM/r_{2}

Therefore,

Difference in potential energy, V = V(r_{2}) – V(r_{1}) =-GmM (1/r_{2} – 1/r_{1})

Hence, this formula is more accurate than the formula mg (r_{2}– r_{1}).

** Problem**:- ( Class 11 Physics Gravitation )

Two earth satellites A and B each of mass m are to be launched into circular orbits earth’s surface at altitudes 6400km and 1.92X10^{4} km resp. The radius of the Earth is 6400km.Find (a) The ratio of their potential energies and (b) the ratio of their kinetic energies. Which one has greater total energy?

** Answer**:-

- m
_{a}= mass of satellite A

m_{b}=mass of satellite B

h_{a}=6400km, h_{b}=1.92X10^{4} km

R_{e}=6400km

Potential Energy = -GM_{e}m/ (R_{e}+h)

For A (P.E)_{A} = -GM_{e}m/ (6400+6400)

=-GM_{e}m/12800 —(1)

For B(P.E)_{B} = -GM_{e}m/(6400+1.92X10^{4})

-GM_{e}m/ (6400+19200) —(2)

Divide 1 by 2 we will get

(P.E)_{A /}(P.E)_{B} = 2 : 1

- (K.E)
_{A}= GMm/2×12800 (3)

(K.E)_{B}= GMm/2(1.92 X10^{4 }+6400) (4)

Dividing (3) by (4)

(K.E)_{A}/(K.E)_{B }= GMm/(12800) x 2(1.92 X10^{4 }+6400)/ GMm

(K.E)_{A}/ (K.E)_{B }= 2:1

- Total Energy of A = – GMm/2r r=12800km

Total Energy of B = – GMm/2r r=(1.92×10^{4} +6400)km

Total energy of B is greater than A.

** Problem**: – A rocket is fired vertically with a speed of 5 km s

^{–1}from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10

^{24}kg; mean radius of the earth = 6.4 × 10

^{6}m; G= 6.67 × 10

^{–11}N m

^{2}kg

^{–2}.

** Answer: **Velocity of the rocket, v = 5 km/s = 5 × 10

^{3}m/s

Mass of the Earth, M_{e} = 6.0 × 10^{24} kg

Radius of the Earth, R_{e}=6.4 × 10^{6}m

Height reached by rocket mass, m = h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

= 1/2mv^{2}+(-GM_{e}m/R_{e})

At highest point h,

v=0

And Potential Energy = – (GM_{e}m/R_{e}+h)

Total energy of the rocket

=0+ – (GM_{e}m/R_{e}+h)

=-(GM_{e}m/R_{e}+h)

Total energy of the rocket

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface = Total energy at height h.

1/2mv^{2}+(-GM_{e}m/R_{e}) = – GM_{e}m/R_{e}+h

1/2v^{2} = GM_{e}(1/ R_{e} – 1/ R_{e}+h)

By calculating

1/2v^{2} = gR_{e}h/R_{e}+h

Where g=GM/R_{e}^{2} = 9.8m/s^{2}

Therefore,

v^{2}(R_{e}+h) = (2gR_{e}H)

v^{2}R_{e}=h (2gR_{e}-v^{2})

h=R_{e}-v^{2}/ (2gR_{e}-v^{2})

=6.4 × 10^{6}x(5×10^{3})^{2}/2×9.8×6.4×10^{6}-(5×10^{3})^{2}

h=1.6×10^{6}m

Height achieved by the rocket with respect to the centre of the Earth

=R_{e}+h

=6.4 × 10^{6}x1.6×10^{6}

=8.0×10^{6}m

The distance of the rocket is 8 × 10^{6} m from the centre of the Earth.

** Problem: –** Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid-point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

__Answer:__

- 0;
- –2.7 × 10
^{–8}J /kg; - Yes;
- Unstable

Explanation:-

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

** Problem:-** ( Class 11 Physics Gravitation )

Two stars each of one solar mass (= 2× 10^{30} kg) are approaching each other for a head on collision. When they are a distance 10^{9} km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10^{4} km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

__Answer:-__

Mass of each star, M = 2 × 10^{30} kg

Radius of each star, R = 10^{4} km = 10^{7} m

Distance between the stars, r = 10^{9} km = 10^{12}m

For negligible speeds, v = 0 total energy of two stars separated at distance r

=-GMM/r + 1/2mv^{2}

=-GMM/r + 0 (i)

Now, consider the case when the stars are about to collide:

Velocity of the stars = v

Distance between the centres of the stars = 2R

Total kinetic energy of both stars = 1/2Mv^{2} + 1/2Mv^{2} = Mv^{2}

Total potential energy of both stars =-GMM/2R

Total energy of the two stars =Mv^{2} – GMM/2R (ii)

Using the law of conservation of energy, we can write:

Mv^{2} – GMM/2R =- GMM/r

v^{2}=-GM /r + GM/2R = GM (-1/r + 1/2R)

=6.67×10^{-11} x 2 x10^{30}(-1/10^{12} + 1/2×10^{7})

=13.34×10^{19}(-10^{-12}+5×10^{-8})

=13.34×10^{19}x 5 x 10^{-8}

=6.67×10^{12}

v=√6.67×10^{12}= 2.58×10^{6}m/s

** Problem:- **A 400kg satellite is in circular orbit of radius 2R

_{E}about the Earth. How much energy is required to transfer it to circular orbit of radius 4R

_{E}?What are the changes in the kinetic and potential energies?

__Answer: __

E_{i}= – GM_{e}m/2R_{E}

E_{f }= – GM_{e}m/4R_{E}

ΔE = E_{f} – E_{i}

=- GM_{e}m/2R_{E}(1/4-1/2)

ΔE =GM_{e}m/8 R_{E}

In terms of ‘g’

ΔE = gm R_{E}/8

By putting the values and calculating

ΔE = 3.13 x 10^{9}J

The energy which is required to transfer the satellite to circular orbit of radius 4R_{E} is 3.13 x 10^{9}J.

Change in Kinetic energy Δk=k_{f} – k_{i}

Δk = 3.13 x 10^{9}J

Change in Potential energy ΔV= 2xΔE = -6.25x109J