Question

A satellite of mass 5000 kg orbits the Earth (mass = 6.0 x 1024 kg) and has a period of 6000 s. In the above problem the altitude of the satellite above the Earth's surface is

Answer 1

Answer:

ht = 776.63 km above earth's surface

Explanation:

Since the satellite is moving in a circular path, we know that:

$V = \frac{2\pi *R}{T}$     Let this be eq1

If now we express the sum of forces on the satellite:

$Fg = \frac{K*m_t*m_s}{R^2}=m_s*\frac{V^2}{R}$   Replacing the value from eq1 into this equation:

$\frac{K*m_t*m_s}{R^2}=m_s*\frac{(\frac{2\pi*R }{T} )^2}{R}$  Solving for R:

$R = \sqrt[3]{\frac{K*m_t*T^2}{(2\pi )^2} } =7.15*10^6m$   If we subract the earth radius from this value, we'll get the altitude above earth's surface:

ht = 776.63 km