
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
534k+ 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 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

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

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

