Chapter 5, Problem 49P

To determine

The satellite which orbits the Earth faster than the other.

Answer to Problem 49P

Solution:

The speed of thefirst satellite is 1.24 timesthan that of the second satellite

Explanation of Solution

Given:

The orbiting radius of the first satellite, $r_1=7500\text{km}$

The orbiting radius of the second satellite, $r_2=15000\text{km}$

Formula Used:

Gravitational force:

  $F=G\frac{Mm}{r^2}$

Where, $G$ is the gravitational constant.

  $M$ and m are the masses.

r is the distance between M and m .

Centripetal force:

  $F=\frac{m{v}^2}{r}$

Where, m is the mass, v is the velocity and ris the radius.

Calculation:

  $F=G\frac{Mm}{r^2}=m\frac{v^2}{r}$

  $v^2=G\frac{M}{r}$

  $v=\sqrt{G\frac{M}{r}}$

Where $r=R+r_1$ for the first satellite and $r=R+r_2$ for the second satellite.

Substitute the values:

  $v_1=\sqrt{G\frac{M}{r_1}}$

  $\Rightarrow v_1=\sqrt{G\frac{M}{\left(7500+6371\right)}}$

  $v_2=\sqrt{G\frac{M}{r_2}}$

  $\Rightarrow v_2=\sqrt{G\frac{M}{\left(15000+6371\right)}}$

The ratio of the speeds of the first and the second satellite is:

  $\frac{v_1}{v_2}=\sqrt{\frac{G\frac{M}{\left(7500+6371\right)}}{G\frac{M}{\left(15000+6371\right)}}}$

  $\Rightarrow \frac{v_1}{v_2}=1.24$

Conclusion:

Therefore, the speed of thefirst satellite is 1.24 timesfaster than the second satellite.