Chapter 2, Problem 8C

To determine

(a)

The gravitational force on a satellite of mass 200-kg, orbiting the earth in a circular orbit of radius $4.23\times {10}^7\text{m}$.

Answer to Problem 8C

The gravitational force on a satellite of mass 200-kg, orbiting the earth in a circular orbit of radius $4.23\times {10}^7\text{m}$ is 44.5 N.

Explanation of Solution

Given:

Mass of the satellite

$m=200\text{-kg}$

Radius of the satellite’s orbit

$r=4.23\times {10}^7\text{m}$

Mass of the earth (from standard tables)

$M_{\oplus }=5.972\times {10}^{24}\text{kg}$

Universal gravitational constant

$G=6.67\times {10}^{-11}{\text{N-m}}^2{\text{/kg}}^2$.

Formula used:

The gravitational force between the satellite and the Earth is given by,

$F=\frac{G{M}_{\oplus }m}{r^2}$.

Calculation:

Substitute the values of the variables in the formula and calculate the gravitational force.

$\begin{array}{c}F=\frac{G{M}_{\oplus }m}{r^2}\\ =\frac{\left(6.67\times {10}^{-11}{\text{N-m}}^2{\text{/kg}}^2\right)\left(5.972\times {10}^{24}\text{kg}\right)\left(200\text{-kg}\right)}{{\left(4.23\times {10}^7\text{m}\right)}^2}\\ =44.52\text{ N}\end{array}$.

Conclusion:

The gravitational force on a satellite of mass 200-kg, orbiting the earth in a circular orbit of radius $4.23\times {10}^7\text{m}$ is 44.5 N.

To determine

(b)

The speed of the satellite using the expression for the centripetal force.

Answer to Problem 8C

The speed of the satellite orbiting the Earth with an orbital radius of $4.23\times {10}^7\text{m}$ is $\text{3}\text{.07}\times {10}^3\text{ m/s}$.

Explanation of Solution

Given:

Mass of the satellite

$m=200\text{-kg}$

Radius of the satellite’s orbit

$r=4.23\times {10}^7\text{m}$

Gravitational force between the satellite and the earth $F=44.52\text{ N}$.

Formula used:

The centripetal force needed for the orbital motion of the satellite is provided by the gravitational force. The centripetal force is given by the expression,

$F=\frac{m{v}^2}{r}$.

Calculation:

Rewrite the formula for the speed v of the satellite.

$v=\sqrt{\frac{Fr}{m}}$

Substitute the values of the variables in the formula and calculate the orbital speed of the satellite.

$\begin{array}{c}v=\sqrt{\frac{Fr}{m}}\\ =\sqrt{\frac{\left(44.5\text{ N}\right)\left(4.23\times {10}^7\text{m}\right)}{\left(200\text{-kg}\right)}}\\ =3.069\times {10}^3\text{ m/s}\end{array}$.

Conclusion:

The speed of the satellite orbiting the Earth with an orbital radius of $4.23\times {10}^7\text{m}$ is $\text{3}\text{.07}\times {10}^3\text{ m/s}$.

To determine

(c)

The period of the satellite and show that it has a value equal to 1 day.

Explanation of Solution

Given:

Radius of the satellite’s orbit.

$r=4.23\times {10}^7\text{m}$

The orbital speed of the satellite.

$v=3.069\times {10}^3\text{ m/s}$.

Formula used:

The time period of the satellite is given by the expression,

$T=\frac{2\pi r}{v}$.

Calculation:

Substitute the values of the variables in the formula and calculate the value of the time period of the satellite.

$\begin{array}{c}T=\frac{2\pi r}{v}\\ =\frac{2\left(3.14\right)\left(4.23\times {10}^7\text{m}\right)}{\left(3.069\times {10}^3\text{ m/s}\right)}\\ =8.657\times {10}^4\text{s}\end{array}$

Express the time in days.

$\begin{array}{c}T=\left(8.657\times {10}^4\text{s}\right)\left(\frac{1\text{ h}}{3600\text{ s}}\right)\left(\frac{1\text{ d}}{24\text{ h}}\right)\\ =1.00197\text{ d}\end{array}$.

Conclusion:

Thus, a satellite orbiting the Earth at an orbital radius of $4.23\times {10}^7\text{m}$ is 1 day.