In a satellite if the time of revolution is T, then P E is proportional to

Question

In a satellite if the time of revolution is T, then P E is proportional to (A) T1 / 3 (B) T3 (C) T-2 / 3 (D) T-4 / 3

Tardigrade Admin · Accepted Answer

According to Kepler's third law, T2 propto r3 r= radius of orbit For a satellite of mass m orbiting in an orbit of radius r around planet of mass M, Potential energy ( PE )=(- GMm /r) ⇒ PE propto ( GMm /T2 / 3) ⇒ PE propto T-2 / 3

Q. In a satellite if the time of revolution is $T$ , then $PE$ is proportional to

$T^{1/3}$

$T^{3}$

$T^{- 2/3}$

$T^{- 4/3}$

According to Kepler's third law,
$T^{2} \propto r^{3}$
$r =$ radius of orbit
For a satellite of mass $m$ orbiting in an orbit of radius $r$ around planet of mass $M$ ,
Potential energy $(PE) = \frac{- GM m}{r}$
$\Rightarrow PE \propto \frac{GM m}{T ^{2/3}}$
$\Rightarrow PE \propto T^{- 2/3}$

Q. In a satellite if the time of revolution is T, then PE is proportional to

T1/3

T3

T−2/3

T−4/3