Question

A man weighs $75\;kg$ on the surface of the earth. What will his weight be on a geostationary satellite?

Answer 1

Recall that a geostationary satellite is a satellite that has its rotation synchronized with the rotation of the Earth about its axis. In such a case, think about what the relative velocity between the satellite and the Earth would be. To this end, determine the strength of the acceleration due to Earth’s gravity acting on the satellite moving with such a relative (orbital) velocity, which should subsequently lead you to the appropriate conclusion.

Formula Used:
Orbital speed $v_{orbital} = \sqrt{\dfrac{gR^2}{R+h}}$

Complete Solution:
Let us begin by understanding what a geostationary satellite is.

A geostationary satellite is a satellite that revolves around the earth with the same speed and direction as the rotation of the Earth about its axis (West to East). It appears to be motionless at a fixed point in space when viewed from the Earth’s surface, and the orbit around the Earth about which such a satellite orbits is called the geostationary equatorial orbit (GEO), and is usually located just above the Earth’s equator.

At this position, the speed of the GEO satellite should be about $3\;kms^{-1}$ at an altitude of about $35,786\;km$ from the surface of the Earth. The GEO is used by those satellites that are required to constantly be in the field of one particular place over Earth, like telecommunication satellites and weather monitoring satellites.

Now, we are given that a man weighs $75\;kg$ on the surface of the earth. We are required to reduce his weight if he was on a geostationary satellite.

Recall that the relative position between the earth and a GEO satellite remains fixed throughout the trajectory of the satellite around the Earth. This means that the relative velocity between the GEO satellite and the Earth is zero.

Now, the standard expression for orbital velocity is given as:
$v_{orbital} = \sqrt{\dfrac{gR^2}{R+h}}$, where R is the radius of the Earth and h is the orbit height from the surface of the Earth.
$\Rightarrow v^2_{orbital} = \dfrac{gR^2}{R+h}$

Since the relative velocity between the satellite and the Earth is zero i.e., $v_{orbital}=0$
$\Rightarrow 0 = \dfrac{gR^2}{R+h} = gR^2$
We know that the radius of the Earth $R \neq 0$, this only means that $g=0$.

If the GEO satellite experiences no acceleration due to gravity, this means that the man would experience weightlessness.

Therefore, his weight on the geostationary satellite would be zero.

Note:
Do not get confused between geostationary satellites and geosynchronous satellites.
Though both types of satellites synchronize their orbit with the rotation of the Earth, the geostationary satellites have an orbit that lies on the same plane as the equator of the Earth, whereas Geosynchronous satellites can have a differently inclined plane of orbit around the Earth. This means that the time taken to orbit around the earth is the same for both types of satellites but geostationary satellites (orbiting about the equatorial plane) are just a special case of geosynchronous satellites (satellites rotating synchronously with Earth’s rotation in all different planes around the Earth).