Question

The radii of a planet and its satellite are $2r$ and $r$ and their densities are $\rho$ and $2\rho$ respectively. Their centres are separated by a distance $d$. The minimum speed with which a body should be projected from the mid point of the line joining their centres so that the body escapes to infinity is: ($G$-universal gravitational constant)

• A. $4\left[\sqrt{\frac{10G\pi r^3\rho }{3d}}\right]$
• B. $\sqrt{\frac{40G\pi r^3\rho }{3d}}$
• C. $2\left[\sqrt{\frac{10G\pi r^3\rho }{d}}\right]$
• D. $\frac{1}{4}\left[\sqrt{\frac{10G\pi r^3\rho }{3d}}\right]$

Answer 1

The correct option is A $4\left[\sqrt{\frac{10G\pi r^3\rho }{3d}}\right]$

Let the mass of the body be $M$

Potential energy of the body due to planet is given by:

$U_p=\frac{-G{M}_{p}M}{r_p}$

$=-\frac{G\rho \frac{4}{3}\pi (2r{)}^{3}M}{d/2}$

$=-\frac{64G\pi r^3\rho M}{3d}$

Potential energy of the body due to satellite is given by:

$U_s=\frac{-G{M}_{s}M}{r_s}$

$=-\frac{G2\rho \frac{4}{3}\pi (r{)}^{3}M}{d/2}$

$=-\frac{16G\pi r^3\rho M}{3d}$

Total potential energy of the body is:

$U=U_p+U_s$

$=-\frac{64G\pi r^3\rho M}{3d}-\frac{16G\pi r^3\rho M}{3d}$

$=-\frac{80G\pi r^3\rho M}{3d}$

For body to escape, its speed should be enough so that its total energy becomes greater than zero.

$E=U+KE\ge 0$

$-\frac{80G\pi r^3\rho M}{3d}+\frac{1}{2}M{v}^2\ge 0$

$v\ge 4\sqrt{\frac{10G\pi r^3\rho }{3d}}$