g‘ = g / [1 + ( h / R_e) ]²
Where,
g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth
For h = R_e / 2
g‘ = g / [(1 + (R_e / 2R_e) ]²
= g / [1 + (1/2) ]²  =  (4/9)g
Weight of a body of mass m at height h is given as:
W‘ = mg
= m × (4/9)g  =  (4/9)mg
= (4/9)W
= (4/9) × 63  =  28 N. 

8.16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Weight of a body of mass m at the Earth’s surface, W = mg = 250 N
Body of mass m is located at depth, d = (1/2)R_e
Where,
R_e = Radius of the Earth
Acceleration due to gravity at depth g (d) is given by the relation:
g‘ = (1 – (d / R_e)g
= [1 – (R_e / 2R_e) ]g  =  (1/2)g
Weight of the body at depth d,
W‘ = mg‘
= m × (1/2)g  =  (1/2) mg  =  (1/2)W
= (1/2) × 250  =  125 N

8.17. 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 × 10⁶ m; G= 6.67 × 10^–11 N m² kg^–2.

Velocity of the rocket, v = 5 km/s = 5 × 10³ m/s
Mass of the Earth, M_e = 6 × 10²⁴ kg
Radius of the Earth, R_e = 6.4 × 10⁶ m
Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= (1/2)mv² + (-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)
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface = Total energy at height h
(1/2)mv² + (-GM_em / R_e) = -GM_em / (R_e + h)
(1/2)v² = GM_e [ (1/R_e) – 1 / (R_e + h) ]
= GM_e[ (R_e + h – R_e)  / R_e(R_e+ h) ]
(1/2)v² = gR_eh / (R_e + h)
Where g = GM / R_e² = 9.8 ms^-2
∴ v² (R_e + h) = 2gR_eh
v²R_e = h(2gR_e – v²)
h = R_ev² / (2gR_e –  v²)
= 6.4 × 10⁶ × (5 × 10³)² / [ 2 × 9.8 × 6.4 × 10⁶ – (5 × 10³)²
h = 1.6 × 10⁶ m
Height achieved by the rocket with respect to the centre of the Earth = R_e + h
= 6.4 × 10⁶ + 1.6 × 10⁶  =  8 × 10⁶ m.

8.18. The escape speed of a projectile on the earth’s surface is 11.2 km s^–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Escape velocity of a projectile from the Earth, v_esc = 11.2 km/s
Projection velocity of the projectile, v_p = 3v_esc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = v_f
Total energy of the projectile on the Earth = (1/2)mv_p² – (1/2)mv_esc²
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = (1/2)mv_f²
From the law of conservation of energy, we have
(1/2)mv_p² – (1/2)mv_esc²  =  (1/2)mv_f²
v_f = ( v_p² – v_esc² )^1/2
= [ (3v_esc)² – v_esc² ]^1/2
= √8 v_esc
= √8 × 11.2  =  31.68 km/s.

8.19. A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 ×10²⁴ kg; radius of the earth = 6.4 ×10⁶ m; G = 6.67 × 10^–11 N m² kg^–2.

Mass of the Earth, M = 6.0 × 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, R_e = 6.4 × 10⁶ m
Universal gravitational constant, G = 6.67 × 10^–11 Nm²kg^–2
Height of the satellite, h = 400 km = 4 × 10⁵ m = 0.4 ×10⁶ m
Total energy of the satellite at height h = (1/2)mv² + [ -GM_em / (R_e + h) ]
Orbital velocity of the satellite, v = [ GM_e / (R_e + h) ]^1/2
Total energy of height, h = (1/2)GM_em / (R_e + h) – GM_em / (R_e + h)  =  -(1/2)GM_em / (R_e + h)
The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.
Energy required to send the satellite out of its orbit = – (Bound energy)
= (1/2) GM_em / (R_e + h)
= (1/2) × 6.67 × 10^-11 × 6 × 10²⁴ × 200 / (6.4 × 10⁶ + 0.4 × 10⁶)
= 5.9 × 10⁹ J.

8.20. 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).

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 ] + (1/2)mv²
= [ -GMM / r ] + 0  ….(i)
Now, consider the case when the stars are about to collide:
Velocity of the stars = v
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = (1/2) Mv² + (1/2)Mv² = 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 × 2 × 10³⁰ [ (-1/10¹² ) + (1 / 2 × 10⁷) ]
~ 6.67 × 10¹²
v = ( 6.67 × 10¹²)^1/2  =  2.58 × 10⁶ m/s.

8.21. 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 centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium
stable or unstable?

Answer

Grvitational field at the mid-point of the line joining the centres of the two spheres
= GM/(r/2)² (alog negative r) + GM/(r/2) (along r) = 0

Gravitational potential at the midpoint f the line joining the centres of the two spheres is

V = – GM/r/2 + (-GM/r/2) = -4GM/r = -4 × 6.67 × 10^-11 × 100/1.0
= -2.7 × 10^-8 J/Kg

As the effective force on the body placed at mid-point is zero, sso the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its inital position of equilibrium. Hence, the body is in unstable equilibrium.

8.22. As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 10²⁴ kg, radius = 6400 km.

Mass of the Earth, M = 6.0 × 10²⁴ kg
Radius of the Earth, R = 6400 km = 6.4 × 10⁶ m
Height of a geostationary satellite from the surface of the Earth,
h = 36000 km = 3.6 × 10⁷ m
Gravitational potential energy due to Earth’s gravity at height h,
= -GM / (R + h)
= – 6.67 × 10^-11 × 6 × 10²⁴ / (3.6 × 10⁷ + 0.64 × 10⁷)
= -9.4 × 10⁶ J/kg.

8.23. A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2 × 10³⁰ kg).

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.
Gravitational force, f_g = – GMm / R²
Where,
M = Mass of the star = 2.5 × 2 × 10³⁰ = 5 × 10³⁰ kg
m = Mass of the body
R = Radius of the star = 12 km = 1.2 ×10⁴ m
∴ f_g = 6.67 × 10^-11 × 5 × 10³⁰ × m / (1.2 × 10⁴)²  =  2.31 × 10¹¹ m N
Centrifugal force, f_c = mrω²
ω = Angular speed = 2πν
ν = Angular frequency = 1.2 rev s^–1
f_c = mR (2πν)²
= m × (1.2 ×10⁴) × 4 × (3.14)² × (1.2)² = 1.7 ×10⁵m N
Since f_g > f_c, the body will remain stuck to the surface of the star.

8.24. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10⁸kg; G= 6.67 × 10^–11 m²kg^–2.

Mass of the spaceship, m_s = 1000 kg
Mass of the Sun, M = 2 × 10³⁰ kg
Mass of Mars, m_m = 6.4 × 10 ²³ kg
Orbital radius of Mars, R = 2.28 × 10⁸ kg =2.28 × 10¹¹m
Radius of Mars, r = 3395 km = 3.395 × 10⁶ m
Universal gravitational constant, G = 6.67 × 10^–11 m²kg^–2
Potential energy of the spaceship due to the gravitational attraction of the Sun = -GMm_s / R
Potential energy of the spaceship due to the gravitational attraction of Mars = -GM_mm_s / r
Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.
Total energy of the spaceship =  -GMm_s / R –  GM_mm_s / r
= -Gm_s[ (M / R) + (m_m / r) ]
The negative sign indicates that the system is in bound state.
Energy required for launching the spaceship out of the solar system
= – (Total energy of the spaceship)
= Gm_s[ (M / R) + (m_m / r) ]
= 6.67 × 10^-11 × 10³ × [ (2 × 10³⁰ / 2.28 × 10¹¹) + (6.4 × 10²³ / 3.395 × 10⁶ ) ]
= 596.97 × 10⁹  =  6 × 10¹¹ J.

8.25. A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10^-11 N m² kg^–2.

Initial velocity of the rocket, v = 2 km/s = 2 × 10³ m/s
Mass of Mars, M = 6.4 × 10²³ kg
Radius of Mars, R = 3395 km = 3.395 × 10⁶ m
Universal gravitational constant, G = 6.67× 10^–11 N m² kg^–2
Mass of the rocket = m
Initial kinetic energy of the rocket = (1/2)mv²
Initial potential energy of the rocket = -GMm / R
Total initial energy = (1/2)mv²– GMm / R
If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.
Total initial energy available = (80/100) × (1/2) mv² – GMm / R  =  0.4mv² – GMm / R
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h = -GMm / (R + h)

Applying the law of conservation of energy for the rocket, we can write:
0.4mv² – GMm / R  =  -GMm / (R + h)
0.4v² = GM / R  –  GM / (R + h)
= GMh / R(R + h)
(R + h) / h  =  GM / 0.4v²R
R / h  =  ( GM / 0.4v²R )  –  1
h = R / [ (GM / 0.4v²R) – 1 ]
= 0.4R²v² / (GM – 0.4v²R)
= 0.4 × (3.395 × 10⁶)² × (2 × 10³)² / [ 6.67 × 10^-11 × 6.4 × 10²³  –  0.4 × (2 × 10³)² × (3.395 × 10⁶) ]
= 18.442 × 10¹⁸ / [ 42.688 × 10¹²  –  5.432 × 10¹² ]
= 18.442 × 10⁶ / 37.256
= 495 × 10³ m = 495 km.