What is the gravitational field strength at a height h above the surface of the Earth? R is the radius of the earth

Question

anonymous · Answer

What is the orbital radius and speed of a synchronous satellite which orbits the earth once every 24h? Take G=66710−11Nm2kg2, Mass of the earth is 5981024kg

anonymous · Answer

The gravitational field strength of a fixed mass $m$, $h$ meters above the Earth's surface is:

$$F(h) = G\frac{mM}{(R+h)^2}$$


where $M$ is the mass of the Earth (in $kg$) and $R$ is its average radius (in meters).

When dealing with a very large body of mass $M$ (as the Earth) and another object $m $ is orbiting it, it means that the gravitational force between them is providing the centripetal force that makes the object $m$ follow a circular path around $M$:


$\large{F_{centripetal} = F_{gravitational}}$

$\large{ma_{c} = G\frac{mM}{r^2}}$

$\large{mr\omega^2 = G\frac{mM}{r^2}}$

$\large{\omega^2 = G\frac{M}{r^3}}$

Where $\omega$ is the angular velocity. The angular velocity is the time it takes the satellite to complete one revolution: $\frac{2\pi}{T}$, $T$ is the time it takes to do so. Remember it must be expressed in $rad/s$.

With this information you can solve for the orbital radius $r$ and then find its tangential velocity by using the relationship between $v$ and $\omega$.