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Physics Question on Gravitational Potential Energy

If mass is written as m=kcPG−1/2h1/2m = k c^P G^{-1/2} h^{1/2}, then the value of PP will be:

A

13\frac{1}{3}

B

12\frac{1}{2}

C

2

D

−13-\frac{1}{3}

Answer

12\frac{1}{2}

Explanation

Solution

The given equation is:

m=kcPG−1/2h1/2,m = k c^P G^{-1/2} h^{1/2},

where kk is a dimensionless constant, cc is the speed of light ([c]=[L][T]−1[c] = [L][T]^{-1}), GG is the gravitational constant ([G]=[M]−1[L]3[T]−2[G] = [M]^{-1}[L]^3[T]^{-2}), hh is Planck's constant ([h]=[M][L]2[T]−1[h] = [M][L]^2[T]^{-1}).

The dimensions of mass are:

[m]=[M].[m] = [M].

The dimensions of each term in the equation are:

[cP]=([L][T]−1)P=[L]P[T]−P,[c^P] = ([L][T]^{-1})^P = [L]^P[T]^{-P},

[G−1/2]=([M]−1[L]3[T]−2)−1/2=[M]1/2[L]−3/2[T],[G^{-1/2}] = ([M]^{-1}[L]^3[T]^{-2})^{-1/2} = [M]^{1/2}[L]^{-3/2}[T],

[h1/2]=([M][L]2[T]−1)1/2=[M]1/2[L][T]−1/2.[h^{1/2}] = ([M][L]^2[T]^{-1})^{1/2} = [M]^{1/2}[L][T]^{-1/2}.

Substituting the dimensions into the equation:

[M]=k⋅[L]P[T]−P⋅[M]1/2[L]−3/2[T]−1⋅[M]1/2[L][T]−1/2.[M] = k \cdot [L]^P[T]^{-P} \cdot [M]^{1/2}[L]^{-3/2}[T]^{-1} \cdot [M]^{1/2}[L][T]^{-1/2}.

Combine the dimensions of each term:

[M]=[M]1/2+1/2[L]P−3/2+1[T]−P+1−1/2.[M] = [M]^{1/2 + 1/2}[L]^{P - 3/2 + 1}[T]^{-P + 1 - 1/2}.

Equating powers of each dimension:

For mass [M][M]:

1=12+12.1 = \frac{1}{2} + \frac{1}{2}.

For length [L][L]:

0=P−32+1.0 = P - \frac{3}{2} + 1.

Simplify:

P=12.P = \frac{1}{2}.

For time [T][T]:

0=−P+1−12.0 = -P + 1 - \frac{1}{2}.

Simplify:

P=12.P = \frac{1}{2}.

Therefore, the value of PP is:

12\frac{1}{2}