Question

A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is:
A. $\dfrac{\sqrt{3}}{4}mgR$
B. $\dfrac{1}{2}mgR$
C. $\dfrac{1}{4}mgR$
D. $\dfrac{3}{4}mgR$

Answer 1

Hint: We know that the mass of the satellite is given. We have to write two formulas one according to the Newton laws of gravitation and another according to Kepler’s third law. Then we have to apply the law of conservation of energy to get the final minimum value at which the satellite will launch.

Formula used:
$g=\dfrac{GM}{{{R}^{2}}}$
$v=\sqrt{\dfrac{GM}{r}}$
${{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}$

Complete step-by-step answer:
We know from the question that the satellite is of mass ‘m’.
We further know that from Newton’s laws of gravitation,
$g=\dfrac{GM}{{{R}^{2}}}$ and,
We also know that Kepler's Third Law relates the radius of an orbit to the period of an orbit.
So,
$v=\sqrt{\dfrac{GM}{r}}$
Hence, from the law of energy conservation,
${{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}$,
Gravitational force of the earth here is acting as an opposite force so we are considering it as a negative force,
So,
$-\dfrac{GMm}{R}+{{K}_{f}}=-\dfrac{GMm}{2R}+\dfrac{1}{2}{mv_{0}}^{2}$
Therefore, ${{K}_{i}}=\dfrac{GMm}{2R}+\dfrac{1}{2}m{{(\sqrt{\dfrac{GM}{2R}})}^{2}}$
On solving the above equation we get,
$\Rightarrow {{K}_{i}}=\dfrac{3GMm}{4R}\to {{K}_{i}}=\dfrac{3}{4}mgR$
Therefore option D, is the correct answer.

So, the correct answer is “Option D”.

Additional Information: The magnitude of an object in circular velocity in orbit is equal to the circumference of an orbit divided by the period of the orbit.
According to newton’s law of gravitation, it is stated that each and every particle attracts each and every other particle in the universe with a force that is directly inversely proportional to the square of the distance between the centers and directly proportional to the masses.

Note: In the formula $g=\dfrac{GM}{{{R}^{2}}}$, G is the gravitational constant, M is the mass of the earth and ‘R’ is the radius of the earth. Here in the formula ${{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}$, U is the energy which is opposing the satellite at the initial and final stage and K is the kinetic energy at the final and at the initial stage.