Question

A satellite is to revolve around the earth in a circle of radius 8000 km. With what speed should this satellite be projected into orbit? What will be its period of revolution? ( $g = 9.8\,m/{s^2}$ and radius of the earth =6400 km).

Answer 1

Hint Use the relation of orbital velocity to determine the speed with which the projectile should be projected and substitute the value of gravitational acceleration. The time period can be calculated as the ratio of the circumference of orbit to the orbital velocity.
Formula used:
-Orbital velocity of a satellite ${v_o} = \sqrt {\dfrac{{GM}}{R}}$ where $G$ is the gravitational constant, $M$ is the mass of Earth and $R$ is the distance of the satellite from the Earth
- ${\text{T = }}\dfrac{{{\text{2}}\pi R}}{{{v_o}}}$ where $T$ is the time period of revolution.

Complete step by step answer
The speed with which the satellite should be launched into orbit is equal to the orbital velocity. However since we haven’t been given the mass of Earth, we can use the relation of gravitational acceleration at the surface of the Earth:
$\Rightarrow g = \dfrac{{GM}}{{{r^2}}}$ where $r$ is the radius of Earth
$\Rightarrow GM = g{r^2}$
Substituting the value of $GM$ in the formula of orbital velocity, we get:
$\Rightarrow {v_o} = \sqrt {\dfrac{{g{r^2}}}{R}}$
On substituting the values of $g = 9.8\,m/{s^2}$ , $r = 6400\,kms = 6400 \times {10^3}m$ , $R = 8000\,kms = 8000 \times {10^3}m$ , we get
$\Rightarrow {v_0} = \sqrt {\dfrac{{9.8 \times {{\left( {6400 \times {{10}^3}} \right)}^2}}}{{8000 \times {{10}^3}}}}$
${v_0} = 7084\,m/s = 7.08\,4km/s$ which is the velocity required to launch the satellite into orbit.
Since the orbit is circular in nature, the circumference of orbit $= 2\pi R$ and we can calculate the time period of one revolution as
$\Rightarrow {\text{T = }}\dfrac{{{\text{2}}\pi R}}{{{v_o}}}$
$\Rightarrow {\text{T = }}\dfrac{{{\text{2}}\pi (8000 \times {{10}^3})}}{{7084}} = 7095s$
which is roughly equal to 118 minutes.

Note
The orbital velocity is derived from the balance of gravitation acceleration of the satellite due to Earth’s gravity and centripetal acceleration due to its revolution. Here, we have assumed that the orbit of the satellite is circular in shape but in reality however, the orbits are not completely circular but elliptical in shape. The fact that we know the gravitational acceleration at the surface of the Earth must be utilized to eliminate the variable of mass of the Earth from the equations.