Q#10
The time period of an earth satellite in circular orbit is independent of
(a) the mass of the satellite(b) radius of the orbit
(c) none of them
(d) both of them
Answer: (a)
If the radius of the orbit = a, then velocity of an earth satellite
v = √(GM/a) and the time period T = 2πa/v.
As we can see that both v and T are independent of the mass of the satellite but both depend on the radius of the orbit a.
Q#11
The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. the kinetic energy of the moon with respect to earth is K.
(a) U < K (b) U > K (d) U = K
Answer: (b)
Let the mass of the moon = m, velocity v and its distance from the earth = a.
v = √(GM/a)
K.E. = K = ½mv² = ½mGM/a
GMm/a = 2K
The magnitude of gravitational potential energy of the moon
U = GMm/a = 2K
Obviously U > K.
Q#12
Figure (11-Q2) shows the elliptical path of a planet about the sun. The two shaded parts have equal area. If t₁ and t₂ be the time taken by the planet to go from a to b and from c to d respectively,
(a) t₁ < t₂
(b) t₁ = t₂
(c) t₁ > t₂
(d) insufficient information to deduce the relation between t₁ and t₂.
Answer: (b)
It is the second law of the planetary motion in elliptical orbit by Kepler that the radius vector from Sun to the planet sweeps out equal area in equal time.
Q#13
A person sitting in a chair in a satellite feels weightless because
(a) the earth does not attract the object in a satellite
(b) the normal force by the chair on the person balances the earth's attraction
(c) the normal force is zero
(d) the person in a satellite is not accelerated
Answer: (c)
Only (c) is correct. The weight and the centrifugal force balance each other, hence the normal force is zero.
Q#14
A body is suspended from a spring balance kept in a satellite. The reading of the balance is W₁ when the satellite goes in an orbit of radius R and is W₂ when it goes in an orbit of radius 2R.
(a) W₁ = W₂
(b) W₁ < W₂
(c) W₁ > W₂
(d) W₁ ≠ W₂
Answer: (a)
The weight of a body in a satellite in an orbit is zero whatever the radius of the orbit may be. If the spring balance shows a reading W₁ then it must be due to rotation of the satellite about its own axis. Assuming this rotation same in another orbit, W₁ = W₂. See the diagram below,
The kinetic energy needed to project a body of mass m from the earth's surface to infinity is
(a) ¼mgR
(b) ½mgR
(c) mgR
(d) 2mgR
Answer: (c)
The kinetic energy should be equal to the gravitational potential energy K=U = GMm/R.
Since g = GM/R²
GM =gR²
Hence, K = gR²m/R = mgR
Q#16
A particle is kept at rest at a distance R (earth's radius) above the earth's surface. The minimum speed with which it should be projected so that it does not return is
(a) √(GM/4R)
(b) √(GM/2R)
(c) √(GM/R)
(d) √(2GM/R)
Answer: (c)
The particle at rest is at 2R away from the earth's center. So, Escape velocity
u = √(2GM/2R) = √(GM/R)
Q#17
A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is vₑ. It's speed with respect to the satellite
(a) will be less than vₑ
(b) will be more than vₑ
(c) will be equal to vₑ
(d) will depend on the direction of projection.
Answer: (d)
Its minimum escape speed will be vₑ when projected vertically upward, but if projected otherwise the speed will be greater. So, the escape speed will depend on the direction of projection.
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