
A uniform spherical shell gradually shrinks maintaining its shape. The gravitational potential at the center:
A. Increases
B. Decreases
C. Remains constant
D. Oscillates
Answer
496.5k+ views
Hint: Similar to electrostatic potential energy, gravitational potential energy is the work done in bringing the object from infinity to some point in a gravitational field. It is assumed that the gravitational potential energy is zero at infinity. Work is done on the object to bring it from infinity, this work is stored in the form of gravitational potential energy.
Complete step by step solution:
The gravitational potential of a spherical shell of mass $m$and radius $r$ is given below.
${V_G} = - \dfrac{{mg}}{r}$
Here, $g$ is acceleration due to gravity.
Now, according to the question, the spherical shell gradually shrinks maintaining its shape. It means the radius of the spherical shell decreases.
From the formula, we see that, if the radius decreases then the quantity on the right-hand side increases.
But due to the presence of a negative sign on the right-hand side, the potential decreases.
$r\left( \downarrow \right) \Rightarrow \dfrac{{mg}}{r}\left( \uparrow \right) \Rightarrow {V_g}\left( \downarrow \right)$
Therefore, we can say that when the radius decreases, the potential at the center also decreases.
Hence, the correct option is (B) decreases.
Note:
Objects having masses apply attractive force on each other called gravitational force and the space around the mass object is called its gravitational field.
The gravitational strength is defined as the gravitational force experienced by the unit mass object.
The unit of gravitational potential is joule per kilogram and the dimension is $\left[ {{L^2}{T^{ - 2}}} \right]$.
Gravitational potential energy is always negative because work is obtained. It is a scalar quantity.
Complete step by step solution:
The gravitational potential of a spherical shell of mass $m$and radius $r$ is given below.
${V_G} = - \dfrac{{mg}}{r}$
Here, $g$ is acceleration due to gravity.
Now, according to the question, the spherical shell gradually shrinks maintaining its shape. It means the radius of the spherical shell decreases.
From the formula, we see that, if the radius decreases then the quantity on the right-hand side increases.
But due to the presence of a negative sign on the right-hand side, the potential decreases.
$r\left( \downarrow \right) \Rightarrow \dfrac{{mg}}{r}\left( \uparrow \right) \Rightarrow {V_g}\left( \downarrow \right)$
Therefore, we can say that when the radius decreases, the potential at the center also decreases.
Hence, the correct option is (B) decreases.
Note:
Objects having masses apply attractive force on each other called gravitational force and the space around the mass object is called its gravitational field.
The gravitational strength is defined as the gravitational force experienced by the unit mass object.
The unit of gravitational potential is joule per kilogram and the dimension is $\left[ {{L^2}{T^{ - 2}}} \right]$.
Gravitational potential energy is always negative because work is obtained. It is a scalar quantity.
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