Question

check the correctness of the relation escape velocity, v=√2GM/R

Answer 1

The dimensional correctness of the relation escape velocity, v=√2GM/R is LT^-1 = LT^-1.

• Given,
• $v=\sqrt{\dfrac{2GM}{R} }$
• where,
• v = escape velocity = ms^-1
• G = gravitational constant = Nm^2 / Kg^2
• M = mass = Kg
• R = radius = m
• we have,
• v = ms^-1 = [LT^-1]
• G = Nm^2 / Kg^2 = [M^-1L^3T^-2]
• M = Kg = [M]
• R = m = [L]
• now,
• $[LT^{-1}] = \sqrt{\dfrac{[M^{-1}L^3T^{-2}][M]}{[L]} } \\\\=\sqrt{L^2T^{-2}} \\\\=[LT^{-1}]$
• Therefore,
• LHS=RHS.

Answer 2

Answer:

The given formula $v = \sqrt{\frac{2GM}{R} }$ is correct.

Explanation:

Let us compare the dimensional formula of $L.HS$ and $R.H.S$.

In the case of $L.H.S$ :

$v$ is the escape velocity.

Hence, its unit is $ms^{-1}$.

Therefore, its dimensions will be $[M^{0} L^{1} T^{-2} ]$.

In the case of $R.H.S$ :

In $\sqrt{\frac{2GM}{R} }$, $G$ is the gravitational constant having unit $\frac{Nm^{2} }{kg^{2} }$ and dimension$[M^{-1} L^{3} T^{-2} ]$

M is the mass having unit $kg$ and dimension $[M]$

R is the radius having unit $m$ and dimension $[L]$

Hence, the dimensional formula becomes,

$\sqrt{ \frac{[M^{-1} L^{3}T^{-2} ][M]}{[L]}$

Now, comparing $L.H.S$ and $R.H.S$ we get

$[M^{0} L^{1} T^{-2} ]$ = $\sqrt{ \frac{[M^{-1} L^{3}T^{-2} ][M]}{[L]}$

$[LT^{-1} ]= \sqrt{L^{2} T^{-2} }$

$[LT^{-1}]=[LT^{-1}]$

Therefore, $L.H.S. = R.H.S.$

Hence, the given formula $v = \sqrt{\frac{2GM}{R} }$ is correct.

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