Question

Assuming escape velocity depends upon gravitational constant, radius R of planet, and also its density ρ, derive formula for escape velocity using dimensional analysis.

Answer 1

Hey dear,

◆ Answer -

ve = k.R√Gρ

● Explaination -

Let's first write down dimensions of given quantities -

[ve] = [L1M0T-1]

[G] = [L3M-1T−2]

[R] = [L1M0T0]

[ρ] = [L-3M1T0]

Now, Let's calculate formula for escape velocity taking k as dimensionless constant.

ve = k G^x R^y ρ^z

[L1M0T-1] = [L3M-1T−2]^x [L1M0T0]^y [L-3M1T0]^z

[L1M0T-1] = [L^(3x+y-3z) M^(-x+z) T^(-2x)]

Comparing indexes on two sides -

3x + y - 3z = 1

-x + z = 0

-2x = -1

Solving these three equation,

x = 1/2

y = 1

z = 1/2

Therefore, dimensional formula for escape velocity is -

ve = k G^½ R^1 ρ^½

ve = k.R√Gρ

Thanks for asking..