Question

(a)Show that if a satellite orbits very near the surface of a planet with period \({\bf{T}}\), the density (= mass per unit volume) of the planet is \({\bf{\rho = m/V = }}\frac{{{\bf{3\pi }}}}{{{\bf{G}}{{\bf{T}}^{\bf{2}}}}}\) (b) Estimate the density of the Earth, given that a satellite near the surface orbits with a period of \({\bf{85}}{\rm{ }}{\bf{min}}\). Approximate the Earth as a uniform sphere.

Short answer

(a) The density mass per unit volume is\(\frac{{3\pi }}{{G{T^2}}}\).

(b) The density of the speed of the satellite is \(5432.56\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

Step-by-step solution

Concept

When the satellite rotates around the Earth, it undergoes centripetal force and gravitational acceleration.

The law of conservation of energy is used to calculate the density of the satellite, by equating centripetal force and gravitational force.

The formula for centripetal force is given as,

\({F_c} = \frac{{m{v^2}}}{R}\)

The formula for gravitational acceleration is given as,

\({F_g} = \frac{{GMm}}{{{R^2}}}\)

Using law of conservation of energy to find density.

(a)

The expression for the time period is given as,

\(\begin{aligned}T &= \frac{{2\pi }}{\omega }\\v &= \frac{{2\pi R}}{T}\end{aligned}\)

Here,\(R\)is the radius of the satellite and\(v\)is the velocity of the satellite.

In the space, the satellite observes centripetal as well as gravitational force in order to revolve around the Earth.

By using Conservation of energy, the net force is due to the centripetal force and gravitational force,

\({F_c} = {F_g}\)

Substitute the value of gravitational force and Centripetal force is above equation,

\(\begin{aligned}\frac{{m{v^2}}}{R} &= \frac{{GMm}}{{{R^2}}}\\{\left( {\frac{{2\pi R}}{T}} \right)^2} &= \frac{{GM}}{R}\\M &= \frac{{4{\pi ^2}{R^3}}}{{G{T^2}}}\end{aligned}\)

Here, \(G\) is the gravitational constant, \(M\) is the mass of the planet and \(m\) is the mass of the satellite.

The expression for the density of the planet is given as,

\(\rho = \frac{m}{V}\)

Here,\(V\)is the volume of the satellite.

Substitute the values in the above equation,

\(\begin{aligned}\rho &= \frac{M}{V}\\ &= \frac{{\frac{{4{\pi ^2}{R^3}}}{{G{T^2}}}}}{{\frac{4}{3}\pi {R^3}}}\\ &= \frac{{3\pi }}{{G{T^2}}}\end{aligned}\)

Thus, the density mass per unit volume is \(\frac{{3\pi }}{{G{T^2}}}\).

Calculation of density of the Earth

(b)

Given data:

The time period of the satellite is \(T = 85\;{\rm{min}} = 5100\;{\rm{s}}\).

Standard value:

The radius of Earth is\(R = 6400\;{\rm{km}} = 6.4 \times {10^6}\;{\rm{m}}\).

The expression for the density of the satellite is,

\(\begin{aligned}{\rho _E} &= \frac{{3\pi }}{{G{T^2}}}\\ &= \frac{{3\pi }}{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}} \right){{\left( {85\;\min \times \frac{{60\,{\rm{s}}}}{{1\;\min }}} \right)}^2}}}\\ &= 5432.56\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\end{aligned}\)

Thus, the density of the speed of the satellite is \(5432.56\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).