
Gravitational potential of the body of mass m at a height h from surface of earth of radius R is
$\begin{align}
& a)-g(R+h) \\
& b)-g(R-h) \\
& c)g(R+h) \\
& d)none \\
\end{align}$
Answer
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Hint: Let us find the potential energy of the particle of mass m at a height h given. Next, as acceleration due to gravity is mentioned, let us derive the formula in terms of acceleration due to gravity. Finally, the gravitational potential is the potential energy divided by the mass of the particle.
Formula used:
$U=\dfrac{GMm}{R+h}$
Complete answer:
Let us write the potential energy formula for any object of mass m at height h,
$U=\dfrac{GMm}{R+h}$
As the option are in terms of acceleration due to gravity,
Let us convert G to g,
$\begin{align}
& g=\dfrac{GM}{{{r}^{2}}} \\
& U=\dfrac{-g{{R}^{2}}m}{R(1+\dfrac{h}{R})} \\
\end{align}$
Now, deriving the above equation a step ahead and dividing it with the mass, we will get the gravitational potential,
$\begin{align}
& U=\dfrac{-g{{R}^{2}}m(R-h)}{{{R}^{2}}} \\
& \Rightarrow U=-gm(R-h) \\
& \Rightarrow V=\dfrac{U}{m} \\
& \Rightarrow V=\dfrac{-gm(R-h)}{m} \\
& \Rightarrow V=-g(R-h) \\
\end{align}$
So, the correct answer is “Option B”.
Additional Information:
In classical mechanics the gravitational potential at a location is equal to the work per unit mass that would be needed to move an object to that location from a fixed different location. It is analogous to the electric potential with mass playing the role of charge. In mathematics, the gravitational potential is also known as Newtonian potential and is fundamental in the study of potential theory. It is also used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies. The gravitational potential at a location is the gravitational potential energy buddy unit mass at that location. Potential energy is equal to the work done by a gravitational field moving a body to its given position in space from Infinity.
Note:
The gravitational potential energy will be affected by factors such as its height related to some reference point, it’s Mass, and the strength of the gravitational field it is in. Whereas the gravitational potential does not depend on the mass of the object. It only depends on the height related to some reference point and strength of the gravitational field.
Formula used:
$U=\dfrac{GMm}{R+h}$
Complete answer:
Let us write the potential energy formula for any object of mass m at height h,
$U=\dfrac{GMm}{R+h}$
As the option are in terms of acceleration due to gravity,
Let us convert G to g,
$\begin{align}
& g=\dfrac{GM}{{{r}^{2}}} \\
& U=\dfrac{-g{{R}^{2}}m}{R(1+\dfrac{h}{R})} \\
\end{align}$
Now, deriving the above equation a step ahead and dividing it with the mass, we will get the gravitational potential,
$\begin{align}
& U=\dfrac{-g{{R}^{2}}m(R-h)}{{{R}^{2}}} \\
& \Rightarrow U=-gm(R-h) \\
& \Rightarrow V=\dfrac{U}{m} \\
& \Rightarrow V=\dfrac{-gm(R-h)}{m} \\
& \Rightarrow V=-g(R-h) \\
\end{align}$
So, the correct answer is “Option B”.
Additional Information:
In classical mechanics the gravitational potential at a location is equal to the work per unit mass that would be needed to move an object to that location from a fixed different location. It is analogous to the electric potential with mass playing the role of charge. In mathematics, the gravitational potential is also known as Newtonian potential and is fundamental in the study of potential theory. It is also used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies. The gravitational potential at a location is the gravitational potential energy buddy unit mass at that location. Potential energy is equal to the work done by a gravitational field moving a body to its given position in space from Infinity.
Note:
The gravitational potential energy will be affected by factors such as its height related to some reference point, it’s Mass, and the strength of the gravitational field it is in. Whereas the gravitational potential does not depend on the mass of the object. It only depends on the height related to some reference point and strength of the gravitational field.
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