Question

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 18.0 hours and an orbital radius 2.00 × 10 7 m. The second planet has an orbital radius 4.00 × 10 7 m. What is the period of the second satellite? g = 9.80 m/s^2

Answer 1

Answer:

Time period of second satellite will be 50.904 hour

Explanation:

We have given time period of first satellite $T_1=18hour$

Orbital radius of first satellite $a_1=2\times 10^7m$

Orbital radius of second satellite $a_2=4\times 10^7m$

We have to find the time period of second satellite.

Time period of satellite is equal to $T=2\pi \sqrt{\frac{a^3}{G(M_1+M_2)}}$

From the relation we can see that $T\propto a^{\frac{3}{2}}$

$\frac{T_1}{T_2}=\frac{a_1^\frac{3}{2}}{a_2^\frac{3}{2}}$

$\frac{18}{T_2}=\frac{(2\times 10^7)^\frac{3}{2}}{(4\times 10^7)^\frac{3}{2}}$

$\frac{18}{T_2}=(\frac{1}{2})^\frac{3}{2}$

$T_2=50.904hour$

Time period of second satellite will be 50.904 hour