Answer:
ht = 776.63 km above earth's surface
Explanation:
Since the satellite is moving in a circular path, we know that:
[tex]V = \frac{2\pi *R}{T}[/tex] Let this be eq1
If now we express the sum of forces on the satellite:
[tex]Fg = \frac{K*m_t*m_s}{R^2}=m_s*\frac{V^2}{R}[/tex] Replacing the value from eq1 into this equation:
[tex]\frac{K*m_t*m_s}{R^2}=m_s*\frac{(\frac{2\pi*R }{T} )^2}{R}[/tex] Solving for R:
[tex]R = \sqrt[3]{\frac{K*m_t*T^2}{(2\pi )^2} } =7.15*10^6m[/tex] If we subract the earth radius from this value, we'll get the altitude above earth's surface:
ht = 776.63 km
