
If the gravitational force has varied as \[{{r}^{-5/2}}\]instead of \[{{r}^{-2}}\]; the potential energy of a particle at a distance ‘r’ from the centre of the earth would be proportional to
\[\begin{align}
& A.\,{{r}^{-1}} \\
& B.\,{{r}^{-2}} \\
& C.\,{{r}^{-3/2}} \\
& D.\,{{r}^{-5/2}} \\
\end{align}\]
Answer
520.8k+ views
Hint: We have to consider the formulae of the gravitational force and the gravitational potential energy. Represent the formula of the gravitational potential energy in terms of the gravitational force and equate the change in the value of the radius to obtain the required proportionality value.
Formula used:
\[{{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]
\[U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}\]
Complete answer:
From the given information, we have the data as follows.
The gravitational force has varied as \[{{r}^{-5/2}}\]instead of \[{{r}^{-2}}\].
The formulae that we will be using to solve this problem are as follows.
The gravitational force, \[{{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]
The gravitational potential energy, \[U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}\]
The value of the gravitational potential energy becomes less negative as the radius increases and the value of the gravitational potential energy becomes zero when the radius equals infinite.
Represent the gravitational potential energy in terms of the gravitational force.
\[\begin{align}
& U=\int_{\infty }^{r}{F.\,dr} \\
& \therefore U=\int_{\infty }^{r}{\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}} \\
\end{align}\]
The above equation represents the gravitational potential energy in terms of the gravitational force when the radius is
Replace \[{{r}^{-2}}\]by \[{{r}^{-5/2}}\]
\[\begin{align}
& U=\int_{\infty }^{r}{\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{5/2}}}} \\
& \Rightarrow U=-\dfrac{2}{3}\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{3/2}}} \\
& \therefore U\propto \dfrac{1}{{{r}^{3/2}}} \\
\end{align}\]
Thus, the gravitational potential energy is inversely proportional to the three-two times that of the radius.
\[\therefore \] If the gravitational force has varied as \[{{r}^{-5/2}}\]instead of \[{{r}^{-2}}\]; the potential energy of a particle at a distance ‘r’ from the centre of the earth would be proportional to .
Thus, option (C) is correct.
Note:
The potential energy of the mass increases, when the mass is moved from the surface of the earth to infinite, as the work done will be against the gravitational force. Similarly, the potential energy of the mass decreases, when the mass is moved from infinite to the surface of the earth, as the work done will be along with the gravitational force.
Formula used:
\[{{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]
\[U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}\]
Complete answer:
From the given information, we have the data as follows.
The gravitational force has varied as \[{{r}^{-5/2}}\]instead of \[{{r}^{-2}}\].
The formulae that we will be using to solve this problem are as follows.
The gravitational force, \[{{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]
The gravitational potential energy, \[U=-\dfrac{G{{m}_{1}}{{m}_{2}}}{r}\]
The value of the gravitational potential energy becomes less negative as the radius increases and the value of the gravitational potential energy becomes zero when the radius equals infinite.
Represent the gravitational potential energy in terms of the gravitational force.
\[\begin{align}
& U=\int_{\infty }^{r}{F.\,dr} \\
& \therefore U=\int_{\infty }^{r}{\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}} \\
\end{align}\]
The above equation represents the gravitational potential energy in terms of the gravitational force when the radius is
Replace \[{{r}^{-2}}\]by \[{{r}^{-5/2}}\]
\[\begin{align}
& U=\int_{\infty }^{r}{\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{5/2}}}} \\
& \Rightarrow U=-\dfrac{2}{3}\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{3/2}}} \\
& \therefore U\propto \dfrac{1}{{{r}^{3/2}}} \\
\end{align}\]
Thus, the gravitational potential energy is inversely proportional to the three-two times that of the radius.
\[\therefore \] If the gravitational force has varied as \[{{r}^{-5/2}}\]instead of \[{{r}^{-2}}\]; the potential energy of a particle at a distance ‘r’ from the centre of the earth would be proportional to .
Thus, option (C) is correct.
Note:
The potential energy of the mass increases, when the mass is moved from the surface of the earth to infinite, as the work done will be against the gravitational force. Similarly, the potential energy of the mass decreases, when the mass is moved from infinite to the surface of the earth, as the work done will be along with the gravitational force.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

