Question

A satellite revolves around a planet in circular orbit of radius R (much larger than the radius of the planet) with a time period
of revolution T. Its velocity suddenly becomes zero. (The satellite experiences gravitational force due to the planet only).
(A) It will move along an elliptical path to collide with the planet.
(B) The time of fall of the satellite is nearly
T
18
V2T
(C) The time of fall of the satellite into the planet is nearly
8
(D) It will move along a parabolic path to collide with the planet​

Answer 1

Answer:

A part is the correct answer

Answer 2

Question:

A satellite revolves around a planet in circular orbit of radius R (much larger than the radius of the planet) with a time period of revolution T. Its velocity suddenly becomes zero. (The satellite experiences gravitational force due to the planet only).

(A) It will move along an elliptical path to collide with the planet.

(B) The time of fall of the satellite is nearly $\frac{T}{\sqrt{8} }$

(C) The time of fall of the satellite into the planet is nearly $\frac{\sqrt{2} T }{8}$

(D) It will move along a parabolic path to collide with the planet​

Answer:

The correct answer is option (c)

(C) The time of fall of the satellite into the planet is nearly $\frac{\sqrt{2} T }{8}$

Explanation:

Given:

Radius = R

Radius of circular orbit is much larger than the radius of the planet.

Time period of Revolution  = T

To find:

What happens when the velocity suddenly becomes zero.

Solution:

It will fall because mg is acting towards the center of planet and the initial velocity is zero. It will move in straight line.

By energy conservation:

$\frac{1}{2}m v^{2} = \frac{G M m}{r} = 0 - \frac{G M m}{r}\\ We get V = f(r)\\V = - \frac{dr}{dt}$

$f(r) = - \frac{dr}{dt} \\\int\limits^R_R {- \frac{dr}{f(r)} } \, = - \int\limits^t_0 {} \, dt$

R' = radius of the planet

R' ⇒ 0 [ R >> R']

We get :

$t = \frac{T}{4 \sqrt{2} }\\ Therefore,\\\frac{G M m}{R^{2} } = m (\frac{2 \pi }{T} )^{2} R$

It will fall on the planet and The time of fall of the satellite into the planet is nearly $\frac{\sqrt{2} T }{8}$

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