Name: Escape Velocity: Derivation
Uploaded: 2019-07-19
Duration: 7 min 7 s
Description: Do you know how much velocity is required to escape from Earth to space? In this video, we will discuss the escape velocity of bodies and their values.

Question

A solid sphere of mass $m$ and radius $r$ is placed inside a hollow thin spherical shell of mass $M$ and radius $R$ as shown. A particle of mass $m^{'}$ is placed on the line joining the two centres at a distance $x$ from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if

$(a)$ $r < x < 2 r$

$(b)$ $2 r < x < 2 R$

$(c)$ $x > 2 R$

Accepted Answer

$(a)$ $r < x < 2 r$

Let $A$ be the centre of solid sphere and $Q$ be the centre of hollow sphere.

Let $P$ be the point where the particle of mass $m'$ is placed.

The point is inside the sphere and the shell.

So,

According to the principle of superposition, the magnitude of gravitational force on this particle is the sum of the forces on $m'$ due to solid sphere and hollow sphere.

$\begin{matrix} F_{on m'} & = & F_{solid sphere} & + & F_{hollow sphere} \end{matrix}$

$\begin{matrix} F_{on m^{'}} & = & m' (E_{P}) & + & 0 \\ (E_{P} is gravitational field at P) & (field inside hollow sphere is zero) \end{matrix} \begin{matrix} \end{matrix}$

$F_{on m'} = m' (\frac{G m (x - r)}{r^{3}})$

where,

$m$ is the mass of solid sphere,

$r$ is the radius of solid sphere,

$x$ is the distance from $O$ to $P$ ,

$G$ is the universal gravitational constant.

$F_{on m'} = \frac{G m m' (x - r)}{r^{3}}$

$(b)$ $2 r < x < 2 R$

Let $P$ be the point where $m'$ is placed.

The point is now outside the sphere but inside the shell.

The sphere can now be taken as a point particle located at the centre of the sphere.

According to principle of superposition, the magnitude of gravitational force on this particle is the sum of forces due to hollow sphere and solid sphere.

$\begin{matrix} F_{on m'} & = & F_{solid sphere} & + & F_{hollow sphere} \end{matrix}$

$\begin{matrix} F_{on m^{'}} & = & m' (E_{P}) & + & 0 \\ (E_{P} is gravitational field at P) & (field inside hollow sphere is zero) \end{matrix} \begin{matrix} \end{matrix}$

$F_{on m'} = m' (\frac{G m}{{(x - r)}^{2}})$

$F_{on m'} = \frac{G m m'}{{(x - r)}^{2}}$

$(c)$ $x > 2 R$

Let $P$ be the point where $m'$ is placed.

The point mass is now placed outside both the sphere and shell.

So, we can consider both the sphere and shell to be point objects at their centre.

According to principle of superposition, the magnitude of gravitational force on this particle is the sum of forces due to hollow sphere and solid sphere.

$\begin{matrix} F_{on m'} & = & F_{solid sphere} & + & F_{hollow sphere} \end{matrix}$

$\begin{matrix} F_{on m^{'}} & = & m' (E_{P}) & + & 0 \\ (E_{P} is gravitational field at P) & (field inside hollow sphere is zero) \end{matrix} \begin{matrix} \end{matrix}$

$F_{on m'} = \frac{m' G m}{{(x - r)}^{2}} + \frac{m' G M}{{(x - R)}^{2}}$

where,

$M$ is the mass of hollow sphere,

$R$ is the radius of hollow sphere.

$F_{on m'} = \frac{G m m'}{{(x - r)}^{2}} + \frac{G M m'}{{(x - R)}^{2}}$

Important Questions on Gravitation

Learn and Prepare for any exam you want !