Question

The equation that governs the circular orbit of a satellite is T = 2(pi)√(r3 / Gme) where me is the mass of the earth, which is approximately 6 x 1024kg. Assuming a circular orbit, to achieve an orbit at 20,000km, what would the speed need to be?

Answer 1

`v = 4.47 * 10^3 m/s`

Step-by-step explanation:

As we know that time period is given by the equation

`T = 2\pi \sqrt{(r^3)/(GM)}`

so we know that

`r = 20,000 km = 2* 10^7 m`

`M = 6 * 10^(24) kg`

here we have

`T = 2\pi\sqrt{((2 * 10^7)^3)/((6.67 * 10^(-11))(6 * 10^(24)))}`

`T = 2.81 * 10^4 s`

now in order to find the speed we can say

`speed = (distance)/(time)`

`speed = (2\pi r)/(T)`

`v = (2\pi(2 * 10^7))/(2.81 * 10^4)`

`v = 4.47 * 10^3 m/s`