Question

Gravitational Acceleration inside a Planet

Answer 1

Answer:

PART A

g(R) = $\frac{G(rho)(4)\pi R}{3}$ ;

PART B

g(R) = g(Rp) × $\frac{R}{Rp}$.

Explanation:

Given density of planet = rho.

The planet's radius = Rp.

An object is located a distance R from the center of the planet,

where R< Rp.

The gravitational fore between two point masses m₁ and m₂ is,

F = $\frac{Gm_{1}m_{2}}{r^{2} }$ ; G= universal gravitational constant

                                                                 r = distance between the masses.

for mass m₂ , F=  m₂ g;  where g = acceleration due to gravity;

so, g = $\frac{F}{m_{2} }$ =  $\frac{Gm_{1}}{r^{2} }$;

From figure, only inside part of the planet exerts force and which can be treated as a point mass.

so, g =  $\frac{Gm_{R}}{R^{2} }$

where $m_{R}$ = mass of the planet with radius R.

⇒  $m_{R}$ = rho × $\frac{4}{3}$$×\pi$×R³

 ⇒ g(R) = $\frac{G(rho)(4)\pi R}{3}$ →  PART A

        PART B

  At the surface g(Rp) =  $\frac{G(rho)(4)\pi Rp}{3}$

⇒  g(R) = g(Rp) × $\frac{R}{Rp}$