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(h₂-h₁) (body is brought from position A to B)
• =mgh₂-mgh₁
• W_AB=V_A-V_B
• where
  • V_A=potential energy at point A
  • V_B= potential energy at point B
• 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 = r₁ to r = r₂ (r₂> r₁) along a vertical path,
• We will get , W=V (r₂) – V (r₁)

Conclusion: –

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

V(r) = -GM_em/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/r₂– 1/r₁) is more/less accurate than the formula mg(r₂– r₁) for the difference of potential energy between two points r₂and r₁ 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/r²

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 r₂ and r₁ distance away from the centre of the Earth is respectively given by:

V (r₁) = – G mM/r₁

V (r₂) = -G mM/r₂

Therefore,

Difference in potential energy, V = V(r₂) – V(r₁) =-GmM (1/r₂ – 1/r₁)

Hence, this formula is more accurate than the formula mg (r₂– r₁).

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 1.92X10⁴ 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=1.92X10⁴ km

R_e=6400km

Potential Energy = -GM_em/ (R_e+h)

For A (P.E)_A = -GM_em/ (6400+6400)

            =-GM_em/12800 —(1) 

For B(P.E)_B = -GM_em/(6400+1.92X10⁴)

                        -GM_em/ (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 X10⁴ +6400)   (4)

Dividing (3) by (4)

(K.E)_A/(K.E)_B = GMm/(12800)  x 2(1.92 X10⁴ +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=(1.92×10⁴ +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 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶m;  G= 6.67 × 10^–11 N m² kg^–2.

Answer: Velocity of the rocket, v = 5 km/s = 5 × 10³ m/s

Mass of the Earth, M_e = 6.0 × 10²⁴ kg

Radius of the Earth, R_e=6.4 × 10⁶m

Height reached by rocket mass, m = h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

= 1/2mv²+(-GM_em/R_e)

At highest point h,

v=0

And Potential Energy = – (GM_em/R_e+h)

Total energy of the rocket

=0+ – (GM_em/R_e+h)

=-(GM_em/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.

1/2mv²+(-GM_em/R_e) = – GM_em/R_e+h

1/2v² = GM_e(1/ R_e – 1/ R_e+h)

By calculating

1/2v² = gR_eh/R_e+h

Where g=GM/R_e² = 9.8m/s²

Therefore,

v²(R_e+h) = (2gR_eH)

v²R_e=h (2gR_e-v²)

h=R_e-v²/ (2gR_e-v²)

=6.4 × 10⁶x(5×10³)²/2×9.8×6.4×10⁶-(5×10³)²

h=1.6×10⁶m

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

=R_e+h

=6.4 × 10⁶x1.6×10⁶

=8.0×10⁶m

The distance of the rocket is 8 × 10⁶ 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× 10³⁰ kg) are approaching each other for a head on collision. When they are a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Answer:-

Mass of each star, M = 2 × 10³⁰ kg

Radius of each star, R = 10⁴ km = 10⁷ m

Distance between the stars, r = 10⁹ km = 10¹²m

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

=-GMM/r + 1/2mv²

=-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 = 1/2Mv² + 1/2Mv² = Mv²

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

Total energy of the two stars =Mv² – GMM/2R (ii)

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

Mv² – GMM/2R =- GMM/r

v²=-GM /r + GM/2R = GM (-1/r + 1/2R)

=6.67×10^-11 x 2 x10³⁰(-1/10¹² + 1/2×10⁷)

=13.34×10¹⁹(-10^-12+5×10^-8)

=13.34×10¹⁹x 5 x 10^-8

=6.67×10¹²

v=√6.67×10¹²= 2.58×10⁶m/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_em/2R_E

E_f = – GM_em/4R_E

ΔE = E_f – E_i

 =- GM_em/2R_E(1/4-1/2)

ΔE =GM_em/8 R_E

In terms of ‘g’

ΔE = gm R_E/8

By putting the values and calculating

ΔE = 3.13 x 10⁹J

The energy which is required to transfer the satellite to circular orbit of radius 4R_E is 3.13 x 10⁹J.

Change in Kinetic energy Δk=k_f – k_i

Δk = 3.13 x 10⁹J

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