
The gravitational potential due to earth at infinite distance from it is zero. Let the gravitational potential at a point $P$ be $ - 5{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$.Suppose, we arbitrarily assume the gravitational potential at infinity to be $ + 10{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$, then the gravitational potential at $P$ will be?
A. $ - 5{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$
B. $ + 5{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$
C. $ - 15{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$
D. $ + 15{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$
Answer
576.3k+ views
Hint
Use the concept that the potential gradient remains constant so the potential if increased at one point then the potential at all point increases relatively. We will use work done by external force on unit mass.
Complete step-by-step solution:In this question, as the potential at infinity is increased by the amount of $ + 10{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$, hence the potential at all the other points in relation of this is increased by an equal amount everywhere . Hence, the potential at point $P$ will be given as sum of these two potentials.
The gradient of the potential will remain constant throughout the space.
Case 1:
When the gravitational potential due to earth at infinite distance from it is zero, then work done is calculated as,
${W_{ext}} = {V_P} - {V_\infty }$ ---- (1)
Substitute $ - 5$ for ${V_P}$ and $0$ for ${V_\infty }$ in the equation (1).
$
{W_{ext}} = \left( { - 5 - 0} \right) \\
{W_{ext}} = - 5{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}} \\
$
Case 2:
When the gravitational potential due to earth at infinite to be $ + 10{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$, then the gravitational potential at $P$ is calculated as,
\[{W_{ext}} = {V_P} - {V_\infty }\] ---- (2)
Substitute $ - 5$ for \[{W_{ext}}\] and $ - 10$ for ${V_\infty }$ in the equation (2).
\[
- {\text{5}} = {V_P} - {\text{10}} \\
{V_P} = {\text{5 Jk}}{{\text{g}}^{ - {\text{1}}}} \\
\]
Therefore, the gravitational potential at $P$ will be \[{\text{5 Jk}}{{\text{g}}^{ - {\text{1}}}}\].
Hence, the correct option is (B).
Note:-
Divide the question in two parts and solve carefully. Substitute the values as per given situation and do the calculation carefully. Avoid silly mistakes while forming equations and simplifying the equations.
Use the concept that the potential gradient remains constant so the potential if increased at one point then the potential at all point increases relatively. We will use work done by external force on unit mass.
Complete step-by-step solution:In this question, as the potential at infinity is increased by the amount of $ + 10{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$, hence the potential at all the other points in relation of this is increased by an equal amount everywhere . Hence, the potential at point $P$ will be given as sum of these two potentials.
The gradient of the potential will remain constant throughout the space.
Case 1:
When the gravitational potential due to earth at infinite distance from it is zero, then work done is calculated as,
${W_{ext}} = {V_P} - {V_\infty }$ ---- (1)
Substitute $ - 5$ for ${V_P}$ and $0$ for ${V_\infty }$ in the equation (1).
$
{W_{ext}} = \left( { - 5 - 0} \right) \\
{W_{ext}} = - 5{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}} \\
$
Case 2:
When the gravitational potential due to earth at infinite to be $ + 10{\text{ Jk}}{{\text{g}}^{ - {\text{1}}}}$, then the gravitational potential at $P$ is calculated as,
\[{W_{ext}} = {V_P} - {V_\infty }\] ---- (2)
Substitute $ - 5$ for \[{W_{ext}}\] and $ - 10$ for ${V_\infty }$ in the equation (2).
\[
- {\text{5}} = {V_P} - {\text{10}} \\
{V_P} = {\text{5 Jk}}{{\text{g}}^{ - {\text{1}}}} \\
\]
Therefore, the gravitational potential at $P$ will be \[{\text{5 Jk}}{{\text{g}}^{ - {\text{1}}}}\].
Hence, the correct option is (B).
Note:-
Divide the question in two parts and solve carefully. Substitute the values as per given situation and do the calculation carefully. Avoid silly mistakes while forming equations and simplifying the equations.
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