Question

The Dimensional formula of gravitational potential the same as?
$\left( a \right)$ Work
$\left( b \right)$ Force
$\left( c \right)$ Velocity
$\left( d \right){\left( {{\text{velocity}}} \right)^2}$

Answer 1

Hint – In this particular question use the concept that gravitation energy is directly proportional to the mass of the earth and inversely proportional to the radius of the earth and the proportionality constant is universal gravitational constant so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Now as we know that the gravitational energy is given as,

Gravitational energy (E) = ($\dfrac{{GM}}{R}$)........................... (1)

Where G = universal gravitational constant, M = mass of the Earth and R = radius of the earth.
As we know force between two masses (m) and (M) is

$F = \dfrac{{G\left( {m.M} \right)}}{{{r^2}}}$, where G is called a gravitational constant and r is the distance between them.

So, the formula of universal gravitational constant G is

$\Rightarrow G = \dfrac{{F.{r^2}}}{{m.M}}$

Now as we know force is the product of mass (M) and acceleration (a)

Therefore, F = (M. a).

Now as we know that the dimension of mass (M) is (M1).
And we know the S.I unit of acceleration (a) is m/s2.

The dimension of meter is (L1) and the dimension of second (s) is (T1).
So the dimension of acceleration is (L1 T -2).
Therefore, the dimension of force (F) is [M1 L1 T -2].
And we all know distance is measured in meters so the dimension of r is [L1].
Therefore, the dimension of G is

$\Rightarrow G = \dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {{M^2}} \right]}}$

Now on simplifying we have,
$G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$
Now from equation (1) we have,

$\Rightarrow E = \dfrac{{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]\left[ M \right]}}{{\left[ L \right]}}$
$\Rightarrow E = \left[ {{L^2}{T^{ - 2}}} \right]$............... (2)

So this is the dimension of the gravitational potential.
Now as we know that the unit of velocity is meter/second.

So the dimension of velocity is $\dfrac{{\left[ L \right]}}{{\left[ T \right]}} = \left[ {L{T^{ - 1}}} \right]$
So the dimension of square of velocity is, ${\left[ {L{T^{ - 1}}} \right]^2} = \left[ {{L^2}{T^{ - 2}}} \right]$............ (3)

So from equation (2) and (3) we can say that dimension of gravitational potential = ${\left( {{\text{velocity}}} \right)^2}$
Hence option (D) is the correct answer.

Note – The basics of dimension formulas are very helpful for solving these kinds of problems. Dimensional formula is basically an expression depicting the units of physical quantity in terms of the fundamental quantities that are worldwide used for measurement. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M, L and T.