Chapter 9, Problem 47SP

A satellite orbits the Earth at a height of 200 km in a circle of radius 6570 km. Find the linear speed of the satellite and the time taken to complete one revolution. Assume the Earth’s mass is 6.0 × 1024 kg. [Hint: The gravitational force provides the centripetal force.]

To determine

The speed of the satellite and the time taken by the satellite which is rotating in a circular orbit at an altitude of $200\text{ km}$ given that the radius of the Earth is $6570\text{ km}$ and the mass of the Earth is $6.0\times {10}^{24}\text{ kg}$.

Answer to Problem 47SP

Solution: $7.7\text{ km/s}$ and $88.21\text{ min}$

Explanation of Solution

Given data:

The altitude of satellite from Earth’s surface is $200\text{ km}$.

The radius of the Earth is $6570\text{ km}$.

The mass of the Earth is $6.0\times {10}^{24}\text{ kg}$.

Formula used:

The expression for gravitational force on the satellite is written as,

$F_G=\frac{G{m}_1{m}_2}{R^2}$

Here, $F_G$ is the gravitational force on satellite, and $R$ is the orbital radius of the satellite, $m_1$ is the mass of the Earth, $m_2$ is the mass of the satellite, $G$ is the gravitational constant.

The centrifugal force, which is equal in magnitude of centripetal force but opposite in the direction that is in outwards direction, on the satellite is expressed as,

$F_C=\frac{m_2{v}^2}{R}$

Here, $F_C$ is the centrifugal force on the satellite, and $v$ is the linear velocity of the satellite.

Explanation:

Consider the expression for gravitational force on the satellite.

$F_G=\frac{G{m}_1{m}_2}{R^2}$

Understand that standard value of $G$ is $6.97\times {10}^{-11}{\text{ m}}^3\text{/}\left(\text{kg}\cdot {\text{s}}^2\right)$. Therefore, substitute $6.97\times {10}^{-11}{\text{ m}}^3\text{/}\left(\text{kg}\cdot {\text{s}}^2\right)$ for $G$, $6.0\times {10}^{24}\text{ kg}$ for $m_1$, and $6570\text{ km}$ for $R$.

$\begin{array}{c}{F}_G=\frac{\left[6.67\times {10}^{-11}{\text{ m}}^3\text{/}\left(\text{kg}\cdot {\text{s}}^2\right)\right]\left(6.0\times {10}^{24}\text{ kg}\right)m_2}{{\left(6570\text{ km}\right)}^2}\\ =\frac{\left[6.67\times {10}^{-11}{\text{ m}}^3\text{/}\left(\text{kg}\cdot {\text{s}}^2\right)\right]\left(6.0\times {10}^{24}\text{ kg}\right)m_2}{\left(43164900{\text{ km}}^2\right)\left(\frac{1000000{\text{ m}}^2}{1{\text{ km}}^2}\right)}\\ =9.27m_2\end{array}$

Consider the expression for centrifugal force on satellite.

$F_C=\frac{m_2{v}^2}{R}$

Substitute $6570\text{ km}$ for $R$

$\begin{array}{c}{F}_C=\frac{m_2{v}^2}{6570\text{ km}}\\ =\frac{m_2{v}^2}{6570\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)}\\ =\frac{m_2{v}^2}{6570000}\end{array}$

Understand that for the continuous rotation of the satellite in the orbit, the gravitational forceon the satellite must be balanced by the centrifugal force on the satellite

$W=F_C$

Substitute $\frac{m_2{v}^2}{6570000}$ for $F_C$ and $9.27m_2$ for $W$

$\begin{array}{c}9.27m_2=\frac{m_2{v}^2}{6570000}\\ 9.27=\frac{v^2}{6570000}\\ v^2=\left(6570000\right)\left(9.27\right)\\ v=\sqrt{\left(6570000\right)\left(9.27\right)}\end{array}$

Further solve as,

$\begin{array}{c}v=7804.09\text{ m/s}\\ =7804.09\text{ m/s}\left(\frac{0.001\text{ km}}{1\text{ m}}\right)\\ \approx 7.8\text{ km/s}\end{array}$

The speed of the satellite is $7.8\text{ km/s}$.

Consider the distance covered by the satellite in one revolution

$d=2\pi R$

Here, $d$ is the distance covered by the satellite in one revolution.

Substitute $6570\text{ km}$ for $R$

$\begin{array}{c}d=2\pi \left(6570\text{ km}\right)\\ =41280.5\text{ km}\end{array}$

Consider the formula for speed of the satellite

$\begin{array}{l}v=\frac{d}{t}\\ t=\frac{d}{v}\end{array}$

Substitute $7.8\text{ km/s}$ for $v$ and $41280.5\text{ km}$ for $d$

$\begin{array}{c}t=\frac{41280.5\text{ km}}{7.8\text{ km/s}}\\ =5292.37\text{ s}\\ =5292.37\text{ s}\left(\frac{\frac{1}{60}\text{ min}}{1\text{ s}}\right)\\ =88.21\text{ min}\end{array}$

The time taken by the satellite for one revolution is $88.21\text{ min}$.

Conclusion:

The speed of the satellite is $7.7\text{ km/s}$.

The time taken by the satellite for one revolution is $88.21\text{ min}$.