Question

Two spherically symmetric planets with no atmosphere have the same average density, but planet B has twice the radius of planet A. A small satellite of mass mA has period TA when it orbits planet A in a circular orbit that is just above the surface of the planet. A small satellite of mass mB has period TB when it orbits planet B in a circular orbit that is just above the surface of the planet.

Answer 1

A period of a satellite is the time taken by the satellite to travel round a

body.

The comparison between the periods  $T_B$, and $T_A$ is $\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}$

Reason:

The period, T, of a satellite is given as follows;

$T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }$

Volume of the planet A = $\dfrac{4}{3} \cdot \pi \cdot r^3$

Mass of planet A, $m_A$ = $\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho$

Volume of the planet B = $\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3$

Mass of planet B, $m_B$ = $\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho$

Period of the satellite on planet A, $T_A$, is given as follows;

$T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }$

Period of the satellite on planet B, $T_B$, is given as follows;

$T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }$

Therefore, get;

$\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}$

Therefore, $T_A$ = (2·√2)·$T_B$

$T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }$

The comparison between $T_A$ and  $T_B$ is therefore;

• $\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}$

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brainly.com/question/1448749