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?
Solution:
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 = \(\frac 12 R_e\)
Where,
Re = Radius of the Earth
Acceleration due to gravity at depth g (d) is given by the relation:
Weight of the body at depth d,
W'= mg
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.
Solution:
8 × 106 m from the centre of the Earth
Velocity of the rocket, v = 5 km/s = 5 × 103 m/s
Mass of the Earth, Me = 6.0 x 1024 kg
Radius of the Earth, Re = 6.4 x 106 m
Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= \(\frac 12mv^2 + (\frac {-GM_em}{R_e})\)
At highest point h,
v = 0
And, Potential energy = \((\frac {-GM_em}{R_e+h})\)
Total energy of the rocket = 0 + \((-\frac {-GM_em}{R_e+h})\) = - \((\frac {-GM_em}{R_e+h})\)
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface = Total energy at height h
Where g = \(\frac {GM}{R^2_e}\) = 9.8 m/s2 (Acceleration due to gravity on the Earth's surface)
Height achieved by the rocket with respect to the centre of the Earth
= Re + h
= 6.4 x 106 + 1.6 x 106
= 8.0 x 106 m
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.
Solution:
Escape velocity of a projectile from the Earth, vesc = 11.2 km/s
Projection velocity of the projectile, vp = 3vesc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = vf
Total energy of the projectile on the Earth = \(\frac 12mv^2_p - \frac 12mv^2_{esc}\)
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = \(\frac 12mv^2_f\)
From the law of conservation of energy, we have
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 .
Solution:
Mass of the Earth, M = 6.0 × 1024 kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 106 m
Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2
Height of the satellite, h = 400 km = 4 × 105 m = 0.4 ×106 m
Total energy of the satellite at height h = \(\frac 12mv^2 + (\frac {-GM_em}{R_e+h})\)
Orbital velocity of the satellite, v= \(\sqrt {\frac {GM_e}{R_e+h}}\)
Total energy of height, 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)
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).
Solution:
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
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 = \(\frac 12mv^2 + \frac 12mv^2 = Mv^2\)
Total potential energy of both stars = Mv2 = \(\frac {GMM}{2R}\) ...(ii)
Using the law of conservation of energy, we can write:
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?
Solution:
0;
–2.7 × 10–8 J /kg;
Yes;
Unstable
The situation is represented in the given figure:
Mass of each sphere, M = 100 kg
Separation between the spheres, r = 1m
X is the mid point between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions.
Gravitational potential at point X:
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.
