Question

An artificial satellite circling the Earth completes each orbit in 129 min
(a) Find the altitude of the satellite.

(b) What is the value of g at the location of this satellite?​

Answer 1

(a) $2.09\cdot 10^6 m$ above Earth's surface

The orbital speed of a satellite orbiting the Earth can be found using the equation

$v=\sqrt{\frac{GM}{r}}$

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

r is the radius of the satellite's orbit

The orbital speed can also be rewritten as the ratio between the circumference of the orbit and the orbital period, T:

$\frac{2\pi r}{T}$

where

T = 129 min = 7740 s is the period

Combining the two equations,

$\frac{2\pi r}{T}=\sqrt{\frac{GM}{r}}$

And solving for r,

$\frac{(2\pi)^2 r^2}{T^2}=\frac{GM}{r}\\r=\sqrt[3]{\frac{GMT^2}{(2\pi)^2}}=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(7740)^2}{(2\pi)^2}}=8.46\cdot 10^6 m$

This is, however, the orbital radius: this means we have to subtract the Earth's radius to find the altitude of the satellite, which is

$R=6.37\cdot 10^6 m$

therefore, the altitude of the satellite is

$h=r-R=8.46\cdot 10^6 - 6.37\cdot 10^6 =2.09\cdot 10^6 m$

b) $5.57 m/s^2$

The value of g at the location of the satellite is given by

$g=\frac{GM}{r^2}$

where:

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$r=8.46 \cdot 10^6 m$ is the radius of the satellite's orbit

Substituting into the equation, we find

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(8.46\cdot 10^6)^2}=5.57 m/s^2$