Question

Please HELP!!! I will give brainliest to whoever gives honest help!!! A 500 kg satellite experiences a gravitational force of 3000 N, while moving in a circular orbit around the earth. Determine the radius of the circular orbit. Determine the speed of the satellite. Determine the period of the orbit.

Answer 1

Given that,

Mass of satellite = 500 kg

Gravitational force = 3000 N

We need to calculate the radius of the circular orbit

Using formula of gravitational force

$F_{g}=\dfrac{GMm}{(R+h)^2}$

Where, G = gravitational constant

R = radius of earth

h = radius of the circular orbit

M = mass of earth

m = mass of satellite

Put the value into the formula

$3000=\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}}{(6.4\times10^{6})^2+h^2}$

$h^2=\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}-3000\times(6.4\times10^{6})^2}{3000}$

$h=\sqrt{\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}-3000\times(6.4\times10^{6})^2}{3000}}$

$h=5073460.35\ m$

$h=5.1\times10^{6}\ m$

(II). We need to calculate the speed of the satellite

Using formula of velocity

$v=\sqrt{\dfrac{GM}{r}}$

Put the value into the formula

$v=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{5.1\times10^{6}}}$

$v=8858.36\ m/s$

$v=8.8\times10^{3}\ m/s$

$v=8.8\ km/s$

(III). We need to calculate the period of the orbit

Using formula of  time period

$T=2\pi\sqrt{\dfrac{r^3}{GM}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{(5.1\times10^{6})^3}{6.67\times10^{-11}\times6\times10^{24}}}$

$T=3617.40\ sec$

$T=1.00\ hr$

Hence, (I). The radius of the circular orbit is $5.1\times10^{6}\ m$

(II). The speed of the satellite is 8.8 km/s.

(III). The period of the orbit is 1.00 hr.