Question

A spy satellite is in circular orbit around Earth. It makes one revolution in \(6.00 \mathrm{h}\). (a) How high above Earth's surface is the satellite? (b) What is the satellite's acceleration?

Short answer

Answer: The altitude of the satellite above Earth's surface is 3.594 x 10^7 meters.

Step-by-step solution

1. Identify relevant physical quantities for the satellite

First, list the given quantities and constants that are relevant to this problem. - Time taken for one revolution (period): \(T = 6.00 \hspace{1mm} h = 21600 \hspace{1mm} s\) - Earth's mass: \(M_{e} = 5.97 \times 10^{24} \hspace{1mm} kg\) - Earth's radius: \(R_{e} = 6.38 \times 10^{6} \hspace{1mm} m\) - Gravitational constant: \(G = 6.67408 \times 10^{-11}\hspace{1mm}m^3⋅kg^{−1}⋅s^{−2}\)

2. Calculate the satellite's angular velocity

We can calculate the angular velocity of the satellite by using the formula \(\omega = \frac{2\pi}{T}\), where \(\omega\) is the angular velocity and \(T\) is the period. $$\omega = \frac{2\pi}{T} = \frac{2\pi}{21600} = 2.908 \times 10^{-4} \hspace{1mm} rad/s$$

3. Determine the satellite's orbital radius

Use the centripetal force equation, which is equal to the gravitational force acting on the satellite: \(m_s \omega^2 r = \frac{G M_e m_s}{r^2}\). The mass of the satellite, \(m_s\), cancels out when we solve for \(r\) (the satellite's orbital radius): $$r = \sqrt[\leftroot{-2}\uproot{2}]{\frac{G M_e}{\omega^2}}$$ Now, plug in the values: $$r = \sqrt[\leftroot{-2}\uproot{2}]{\frac{6.67408 \times 10^{-11} m^{3}⋅kg^{−1}⋅s^{−2} \times 5.97 \times 10^{24} kg}{(2.908 \times 10^{-4} rad/s)^2}} = 4.23 \times 10^{7} m$$

4. Calculate the satellite's altitude above the Earth

To get the satellite's altitude above the Earth's surface, subtract the Earth's radius from the satellite's orbital radius: $$h = r - R_e = (4.23 \times 10^7 m) - (6.38 \times 10^6 m) = 3.594 \times 10^7 m$$ This is the satellite's altitude above the Earth.