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

Time period of second satellite will be 50.904 hour

Step-by-step explanation:

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

Orbital radius of first satellite
`a_1=2* 10^7m`

Orbital radius of second satellite
`a_2=4* 10^7m`

We have to find the time period of second satellite.

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

From the relation we can see that
`T\propto a^{(3)/(2)}`

`(T_1)/(T_2)=(a_1^(3)/(2))/(a_2^(3)/(2))`

`(18)/(T_2)=((2* 10^7)^(3)/(2))/((4* 10^7)^(3)/(2))`

`(18)/(T_2)=((1)/(2))^(3)/(2)`

`T_2=50.904hour`

Time period of second satellite will be 50.904 hour