Question

Choose the correct alternative:
A. If the zero potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
B. The energy required to launch an orbiting satellite out of Earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.

Answer 1

Hint: The question should be solved in two parts. Try solving the A part while keeping in mind a simple case for the given situation, i.e., A satellite experiences the gravitational force which is equal to the centripetal force. For solving part B use the findings from Part A and try to find out some kind of relation between the two cases given in Part B (Such as ratio between the two).

Complete answer:
First, let us solve for the Part A:
For solving the above questions, let us assume a general equation,
Height of satellite + Radius of earth = R
Now, let us solve for the statement A
A satellite experiences the gravitational force which is equal to the centripetal force, i.e.,
$\dfrac{GMm}{{{R}^{2}}}=\dfrac{m{{v}^{2}}}{R}$
Also, as we know that
\[U=\dfrac{-GMm}{R}\]
And $KE=\dfrac{1}{2}m{{v}^{2}}$
From the above discussed equations, we can conclude that
$TE=-\dfrac{1}{2}m{{v}^{2}}=-KE$
Therefore, we can say that if the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
Now for solving the Part B:
Let us take a look at the formula used to calculate the energy required by a body to escape in such a way that total energy will be zero
$TE=-\dfrac{1}{2}m{{v}^{2}}=-\dfrac{GMm}{R}$ (As discussed in part A)
Now, the extra energy for the satellite will be
\[{{E}_{1}}=\dfrac{GMm}{2R}\]
Also, For the stationary object,
$TE=U=-\dfrac{GMm}{R}$
So we can see,
$\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{1}{2}$
From the above ratio, we can say that the energy required to launch an orbiting satellite out of earth's gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of earth's influence.

Note:
The Universal Law Of Gravitation was given by the English Physicist, Isaac Newton. This law states that “Any two particle/body attract will each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres”, i.e.,
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Where G (Gravitational Constant) = \[6.6743\times {{10}^{-}}^{11}~{{m}^{3}}~k{{g}^{-}}^{1}~{{s}^{-}}^{2}\]
And ${{m}_{1}}$, ${{m}_{2}}$ and r, are the masses of two bodies and the distance between them respectively.