Chapter 5, Problem 55P

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

To determine

To Show:If a satellite orbits very near to the surface of a planet with period $T$ , the density of the planet $\rho =\frac{m}{V}=\frac{3\pi }{G{T}^2}$ .

Answer to Problem 55P

Solution:

The density of the planet in terms of period and gravitational constant is obtained as $\rho =\frac{3\pi }{G{T}^2}$ .

Explanation of Solution

When a satellite orbits, the centripetal acceleration will be equal to the gravitational acceleration.

Mathematically,

  ${\omega }^{2}r={\left(\frac{2\pi }{T}\right)}^{2}r=\frac{GM}{r^2}$

Rearranging the above,

  $\begin{array}{c}\frac{M}{r^3}={\left(\frac{2\pi }{T}\right)}^2\frac{1}{G}\\ =\frac{4{\pi }^2}{G{T}^2}\end{array}$

Where, $G$ is the Newton’s gravitational constant and is equal to $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}\text{/kg}$

  $M$ is the mass of the central body

  $r$ is the radius of the orbit

The density is the ratio of mass and unit volume. Mathematically,

  $\begin{array}{c}\rho =\frac{m}{V}\\ =\frac{m}{\left(\frac{4}{3}\pi r^3\right)}\\ =\left(\frac{3}{4\pi }\right)\frac{m}{r^3}\end{array}$

  $\frac{m}{r^3}=\frac{4\pi \rho }{3}$

  $\frac{M}{r^3}=\frac{4{\pi }^2}{G{T}^2}$ .

Equating the above two:

  $\frac{4\pi \rho }{3}=\frac{4{\pi }^2}{G{T}^2}$

On rearrangement:

  $\rho =\frac{3\pi }{G{T}^2}$

Hence, the density of the planet in terms of time period and gravitational constant is obtained.

Conclusion:

Therefore, the density of the planet in terms of time period and gravitational constant is obtained as $\rho =\frac{3\pi }{G{T}^2}$ .

To determine

The density of the Earth.

Answer to Problem 55P

Solution:

The density of the planet is $5694.6{\text{kg/m}}^{\text{3}}$ .

Explanation of Solution

Given:

Period, $T=85\min =5100s$

Formula used:

From the previous part, the density of the planet can be obtained by:

  $\rho =\frac{3\pi }{G{T}^2}$

Calculation:

Substituting the values of $G$ and T:

  $\rho =\frac{3\pi }{6.67\times {10}^{-11}\times {5100}^2}$

  $\begin{array}{c}\rho =\frac{3\pi }{6.67\times {10}^{-11}\times {5100}^2}\\ =5694.6{\text{kg/m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the density of the planet is $5694.6{\text{kg/m}}^{\text{3}}$ .