Even gravity wants me

Classical Mechanics Level 2

A satellite is in a circular orbit very close to the surface of a planet of uniform density ρ \rho . What is the period (in hours) of the satellite's orbit around the planet?

Details and Assumptions

  • G = 6.67 × 10 − 11 N ⋅ m 2 / k g 2 G = 6.67 \times {10}^{-11} N\cdot{m}^{2}/{kg}^{2}
  • ρ = 4.5 g c m 3 \rho = 4.5 \frac{g}{{cm}^{3}}


The answer is 1.56.

View wiki
Rajdeep Dhingra by Rajdeep Dhingra , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Rajdeep Dhingra
Oct 29, 2014

answer answer Putting values we get answer

10 Helpful 1 Interesting 0 Brilliant 2 Confused

Way overrated, but great problem.

Jake Lai - 6 years, 2 months ago

Log in to reply

Thanks. It gets overrated when this question has not been done for a long time.

Rajdeep Dhingra - 6 years, 2 months ago
Ashik Shahriar
Jun 18, 2020

here the formula is T=sqrt(3 pie/(density G)) then divide it by 3600s to get the value in hours

2 Helpful 1 Interesting 2 Brilliant 2 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...