Question

1. the earth has a mass of approximately ( 5.97\times10^{24}\text{ kg} ) and a radius of ( 6.37\times10^{6}\text{ m} ).

a. if a satellite orbits the earth at an altitude of ( 409,000\text{ m} ) above the earths surface, what is the tangential velocity of the satellite? (hint: you dont need to know the mass of the satellite)
b. what is the period of the satellites motion in seconds? what is it in hours?

Explanation:

Step1: Calculate the radius of the satellite's orbit

The radius of the Earth \(R = 6.37\times10^{6}\space m\), and the altitude \(h=409000\space m = 4.09\times 10^{5}\space m\).
The radius of the satellite's orbit \(r=R + h=(6.37\times10^{6}+4.09\times10^{5})\space m= 6.779\times10^{6}\space m\)

Step2: Find the tangential velocity

The formula for the tangential velocity \(v\) of a satellite in orbit around a planet of mass \(M\) is \(v=\sqrt{\frac{GM}{r}}\), where \(G = 6.67\times10^{-11}\space N\cdot m^{2}/kg^{2}\) and \(M = 5.97\times10^{24}\space kg\)

$$
LATEXBLOCK0
$$

Step3: Calculate the period in seconds

The formula for the period \(T\) of a satellite is \(T=\frac{2\pi r}{v}\)

$$
LATEXBLOCK1
$$

Step4: Convert the period to hours

Since \(1\space h = 3600\space s\), then \(T=\frac{5554}{3600}\approx1.54\space h\)

Answer:

a. The tangential velocity of the satellite is approximately \(7665\space m/s\)
b. The period of the satellite's motion is approximately \(5554\space s\) or \(1.54\space h\)