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
585.9k+ 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<
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

