Solve Kepler's Law: Find Satellite's Period

[keeholee]

Kepler's Law help!

Use this equation: T^2 of A / R^3 of A = T^2 of B / R^3 of B

The moon has a period of 27.3 days and has a mean distance of 3.9*10^5 km from the center of the earth. Find the period of an artifical satellite that is 7.5*10^3 km from the center of the earth.

How do i solve this?

 

[SArnab]

Kepler's law states that the period of a satellite squared over its radius cubed is always constant.

Therefore, you can just plug the values into the equation and solve for T^2 of (the period of the artificial satellite).

 

[keeholee]

yes i got .075 for the period. but isn't that way too small considering the distance from center of the Earth to artifical satellite is bigger than the mean distance from the center of the Earth to the moon?

 

[SArnab]

The distance to the artificial satellite from the Earth is a whole 10^2 smaller, according to the question you posted. The moon is farther away. .075 days for the period however, seems off.

 

[keeholee]

oh yea Oo.


welll yea it is really off but is that possible? maybe its in Earth's i have to convert it or something? which i don't know how