Respuesta :
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
[tex]F_{g}=\dfrac{GMm}{(R+h)^2}[/tex]
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
[tex]3000=\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}}{(6.4\times10^{6})^2+h^2}[/tex]
[tex]h^2=\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}-3000\times(6.4\times10^{6})^2}{3000}[/tex]
[tex]h=\sqrt{\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}-3000\times(6.4\times10^{6})^2}{3000}}[/tex]
[tex]h=5073460.35\ m[/tex]
[tex]h=5.1\times10^{6}\ m[/tex]
(II). We need to calculate the speed of the satellite
Using formula of velocity
[tex]v=\sqrt{\dfrac{GM}{r}}[/tex]
Put the value into the formula
[tex]v=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{5.1\times10^{6}}}[/tex]
[tex]v=8858.36\ m/s[/tex]
[tex]v=8.8\times10^{3}\ m/s[/tex]
[tex]v=8.8\ km/s[/tex]
(III). We need to calculate the period of the orbit
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{r^3}{GM}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{(5.1\times10^{6})^3}{6.67\times10^{-11}\times6\times10^{24}}}[/tex]
[tex]T=3617.40\ sec[/tex]
[tex]T=1.00\ hr[/tex]
Hence, (I). The radius of the circular orbit is [tex]5.1\times10^{6}\ m[/tex]
(II). The speed of the satellite is 8.8 km/s.
(III). The period of the orbit is 1.00 hr.
