Question

A satellite that revolves around the earth is a circle of radius 8000 km. The speed at which this satellite be projected into an orbit, will be
A. 3 km/s
B. 16 km/s
C. 7.55 km/s
D. 8 km/s

Answer 1

For the satellite to revolve in the orbit without being unaffected by the earth’s gravitational field, it must have orbital velocity at which it should move. Recall the formula for orbital velocity of the satellite. Convert the radius of the orbit from km to meter.

Formula used:
Orbital velocity, ${v_o} = \sqrt {\dfrac{{GM}}{R}}$
Here, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.

Complete step by step answer:
We have given that the radius of the orbit of the satellite is $R = 8000\,{\text{km}}$.When the satellite is revolving in the orbit, it has orbital velocity at which it should move. We have the formula for the orbital velocity of the satellite in the orbit of radius R,
${v_o} = \sqrt {\dfrac{{GM}}{R}}$
Here, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
Substituting $G = 6.67 \times {10^{ - 11}}$, $M = 5.97 \times {10^{24}}\,{\text{kg}}$ and $R = 8 \times {10^6}{\text{km}}$ in the above equation, we get,
${v_o} = \sqrt {\dfrac{{\left( {6.67 \times {{10}^{ - 11}}} \right)\left( {5.97 \times {{10}^{24}}} \right)}}{{8 \times {{10}^6}}}}$
$\Rightarrow {v_o} = \sqrt {\dfrac{{3.982 \times {{10}^{14}}}}{{8 \times {{10}^6}}}}$
$\Rightarrow {v_o} = \sqrt {4.977 \times {{10}^7}}$
$\Rightarrow {v_o} = 7.55 \times {10^3}\,{\text{m/s}}$
$\therefore {v_o} = 7.55\,{\text{km/s}}$
Therefore, the satellite must be projected into the orbit with speed 7.55 km/s.

So, the correct answer is option C.

Note: The orbital velocity of the satellite is independent of the mass of the satellite. The only variable is the radius of the orbit. If the satellite revolves around the earth at very low altitude, its orbital speed must be very high. If the orbital speed is less than the required, the satellite will be attracted towards the surface of the earth due to the gravitational force.