Question

2.
In a satellite if the time of revolution is T, then kinetic energyis proportional to
(2)
TIER
(4) 7-218
7-213
100
11
+
​

Answer 1

1/T(-²/³)is the correct answer

Answer 2

Answer:

$E\propto T^{-2/3}$

Explanation:

Let T is the time of revolution of the satellite. As per the Kepler's third law of motion, the square of the time period is directly proportional to the cube of semi major axis. Mathematically, it is equal to :

$T^2\propto a^3$

a is the semi major axis

The velocity of the satellite is given by :

$v=\sqrt{\dfrac{GM}{r}}$

$v\propto \sqrt{\dfrac{1}{r}}$

So, $E\propto T^{-2/3}$

Hence, this is the required solution.