Question

The velocity of a satellite revolving around the earth at a height of 3R from the earth’s surface is: ($g = 10m{s^{ - 2}}$ and $R = 6400km$= radius of earth)
A: $2\sqrt 2Km{s^{ - 1}}$
B: $4Km{s^{ - 1}}$
C: $4\sqrt 2 Km{s^{ - 1}}$
D: $8Km{s^{ - 1}}$

Answer 1

Hint: The velocity of a satellite can be determined using the concept of centripetal force for circular motion. The velocity of a satellite revolving around the earth is independent of the mass of the satellite and it only depends it’s distance from the earth and mass of earth(constant).

Complete step by step answer:
When a satellite revolves around the earth it is constantly under the application of gravitational force that is attracting it towards the surface of the earth. But it has a velocity in the tangential direction or a direction perpendicular to the gravitational pulling force, which is why it does not come to the ground and instead revolves around the earth. This is because the tangential velocity creates a centrifugal force that balances the gravitational pull.
This centrifugal force is directly proportional to the square of tangential velocity and hence should have a value that can balance the gravitational pull acting at that height.
Now we all know that, $v = \sqrt {\dfrac{{Gm}}{{R + h}}} $ here, G is the gravitational constant, R is the radius of earth, h is the height of the satellite from the surface of the earth and m is mass of the satellite.
So according to the question h=3R;
Hence, $v = \sqrt {\dfrac{{Gm}}{{4R}}} = \dfrac{1}{2}\sqrt {\dfrac{{Gm}}{R}} $ (equation: 1)
Also, by definition we know that;
$g = \dfrac{{Gm}}{{{R^2}}}$ (equation: 2)
Multiplying r to enumerator and denominator in equation: 1, we get;
$v = \dfrac{1}{2}\sqrt {\dfrac{{GmR}}{{{R^2}}}} $
Now from equation 2 we get;
$v = \dfrac{1}{2}\sqrt {Rg} $
Substituting the respective values we get;
$v = \dfrac{1}{2}\sqrt {10 \times 6400 \times {{10}^3}} = 4km{s^{ - 1}}$
Hence, the velocity of the satellite revolving around the earth at a height of 3R from the surface is 4km/secs.
Therefore option B is correct.

Note: It is important to use the given information in the question if one does not know the value of gravitational constant $G$, as we did by manipulating the equation to use the given value of $g$.