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

ht = 776.63 km above earth's surface

Step-by-step explanation:

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

`V = (2\pi *R)/(T)` Let this be eq1

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

`Fg = (K*m_t*m_s)/(R^2)=m_s*(V^2)/(R)` Replacing the value from eq1 into this equation:

`(K*m_t*m_s)/(R^2)=m_s*(((2\pi*R )/(T) )^2)/(R)` Solving for R:

`R = \sqrt[3]{(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