
The radius of the Earth is R and acceleration due to gravity at its surface is ‘g’. If a body of mass ‘m’ is sent to a height $\dfrac{R}{4}$ from the Earth’s surface, the potential energy is:
a) $mg\dfrac{R}{3}$
b)$mg\dfrac{R}{4}$
c)$mg\dfrac{R}{5}$
d)$mg\dfrac{R}{16}$
Answer
521.1k+ views
Hint: The gravitational potential energy of the Earth is a function of the distance from the centre of the Earth. The gravitational potential energy increases from the surface of the Earth as we go upwards. Hence taking the difference of the energy from the Earth’s surface to the height of $\dfrac{R}{4}$ would yield the required answer.
Formula used:
${{U}_{h}}=\dfrac{GMmh}{R(R+h)}$
Complete answer:
Let us say we have an object of mass ‘m’ such that it is taken above at a height ‘h’ with respect to ground. If ‘M’ is the mass of the Earth, ‘R’ is the radius of the Earth and G is the gravitational constant, then the potential energy acquired by the body with respect to Earth is,
${{U}_{h}}=\dfrac{GMmh}{R(R+h)}$
The acceleration due to gravity(g) is given by $g=\dfrac{GM}{{{R}^{2}}}$
Therefore substituting the value of ‘GM’ in the above equation we get,
$\begin{align}
& {{U}_{h}}=\dfrac{GMmh}{R(R+h)} \\
& \Rightarrow {{U}_{h}}=\dfrac{gRmh}{(R+h)} \\
& {{U}_{h}}=\dfrac{mgh}{(1+\dfrac{h}{R})} \\
\end{align}$
In the question it is given that the body is to be taken to a height of $\dfrac{R}{4}$ from the surface of the Earth. Hence from the above expression, the gravitational potential energy is numerically equal to,
$\begin{align}
& {{U}_{h}}=\dfrac{mgh}{(1+\dfrac{h}{R})} \\
& \Rightarrow {{U}_{h}}=\dfrac{mg\dfrac{R}{4}}{\left( 1+\dfrac{R}{4R} \right)} \\
& \Rightarrow {{U}_{h}}=\dfrac{1}{4}\dfrac{mgR}{\left( \dfrac{5}{4} \right)} \\
& \therefore {{U}_{h}}=\dfrac{mgR}{5} \\
\end{align}$
Therefore the correct answer of the above question is option c.
Note:
The above expression for potential energy is very precise. The above expression only holds valid to a particular altitude as the value of ‘g’ changes with altitude. This expression is only valid near the surface of the Earth, where h<
Formula used:
${{U}_{h}}=\dfrac{GMmh}{R(R+h)}$
Complete answer:
Let us say we have an object of mass ‘m’ such that it is taken above at a height ‘h’ with respect to ground. If ‘M’ is the mass of the Earth, ‘R’ is the radius of the Earth and G is the gravitational constant, then the potential energy acquired by the body with respect to Earth is,
${{U}_{h}}=\dfrac{GMmh}{R(R+h)}$
The acceleration due to gravity(g) is given by $g=\dfrac{GM}{{{R}^{2}}}$
Therefore substituting the value of ‘GM’ in the above equation we get,
$\begin{align}
& {{U}_{h}}=\dfrac{GMmh}{R(R+h)} \\
& \Rightarrow {{U}_{h}}=\dfrac{gRmh}{(R+h)} \\
& {{U}_{h}}=\dfrac{mgh}{(1+\dfrac{h}{R})} \\
\end{align}$
In the question it is given that the body is to be taken to a height of $\dfrac{R}{4}$ from the surface of the Earth. Hence from the above expression, the gravitational potential energy is numerically equal to,
$\begin{align}
& {{U}_{h}}=\dfrac{mgh}{(1+\dfrac{h}{R})} \\
& \Rightarrow {{U}_{h}}=\dfrac{mg\dfrac{R}{4}}{\left( 1+\dfrac{R}{4R} \right)} \\
& \Rightarrow {{U}_{h}}=\dfrac{1}{4}\dfrac{mgR}{\left( \dfrac{5}{4} \right)} \\
& \therefore {{U}_{h}}=\dfrac{mgR}{5} \\
\end{align}$
Therefore the correct answer of the above question is option c.
Note:
The above expression for potential energy is very precise. The above expression only holds valid to a particular altitude as the value of ‘g’ changes with altitude. This expression is only valid near the surface of the Earth, where h<
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