Question

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is the speed at which the satellite travels?

Answer 1

Answer: $V=5839.051m/s$  

Explanation:

According to the Third Kepler’s Law of Planetary motion:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$   (1)

Where;:

$T=1.26(10)^{4}s$ is the period of the satellite

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=5.98(10)^{24}kg$ is the mass of the Earth

$a$  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity $V$ is given by:

$V=\sqrt{\frac{GM}{a}}$   (2)

Now, from (1) we can find $a$, in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$   (3)

$a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}$   (4)

$a=11705845.57m$   (5)

Substituting (5) in (2):

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}$   (6)

$V=5839.051m/s$   (7)  This is the speed at which the satellite travels