Answer:
The satellite that is 7500 km above earth's surface is 1.24 times faster than the satellite at 15000 km above earth's surface
Explanation:
Generally the gravitational force acting on each satellite is mathematically represented as
[tex]F = \frac{G * M_e * m_2 }{r^2}[/tex]
Here [tex]M_e[/tex] is the mass of the earth
[tex]r = r_e + R [/tex]
So [tex] r_e [/tex] is the radius of the satellite and R is the radius of earth with value R = 6371 km
This force is also equivalent to the centripetal force acting on each satellite which is mathematically represented as
[tex]F = \frac{mv^2}{ r^2}[/tex]
So
[tex]\frac{mv^2}{ r^2} = \frac{G * m_1 * m_2 }{r^2}[/tex]
=> [tex]v = \sqrt{ \frac{GM_e}{r} }[/tex]
So for [tex]r = 7500+ 6371 = 13871 \ km [/tex]
We have that
[tex]v_1 = \sqrt{ \frac{GM_e}{13871} }[/tex]
and [tex]r = 15000+ 6371 = 21371 \ km [/tex]
We have that
[tex]v_2 = \sqrt{ \frac{GM_e}{21371} }[/tex]
So
[tex]\frac{v_1}{v_2} = \frac{\sqrt{\frac{GM_e}{13871} } }{ \sqrt{\frac{GM_e}{21371}} }[/tex][tex]v = \sqrt{ \frac{GM_e}{r} }[/tex]
=> [tex]\frac{v_1}{v_2} = \sqrt{ \frac{21371}{13871} }[/tex]
=> [tex]v_1 = 1.24 v_2[/tex]
So satellite that is 7500 km above earth's surface is 1.24 times faster than the satellite at 15000 km above earth's surface
