Question

In a satellite if the time of revolution is T, then kinetic energy is proportional to

• A. $\frac { 1 } { T }$
• B. $\frac { 1 } { T ^ { 2 } }$
• C. $\frac { 1 } { T ^ { 3 } }$
• D. $T ^ { - 2 / 3 }$

Answer 1

Answer: (D)

$T ^ { - 2 / 3 }$

Answer 2

Time period $T \propto r ^ { 3 / 2 }$ ⇒ $r \propto T ^ { 2 / 3 }$ and Kinetic energy $\propto \frac { 1 } { r } \propto \frac { 1 } { T ^ { 2 / 3 } } \propto T ^ { - 2 / 3 }$.