Chapter 6.2, Problem 40E

a.

To determine

To find: the linear velocity of a point on each planet’s equator.

Answer to Problem 40E

  $\begin{array}{c}{v}_{Mercury}=10.9km/h\\ v_{Venus}=6.5km/h\\ v_{Earth}=1668.5km/h\\ v_{Mars}=861.2km/h\end{array}$

Explanation of Solution

Given information:

The time of rotation $\left(t\right)$ and radius $\left(r\right)$ of first four planets of solar system is shown below.

Planet	Radius (in kilometres)	Time for one rotation (hours)
Mercury	2440	1407.6
Venus	6052	5832.5
Earth	6356	23.935
Mars	3375	24.623

Calculation:

The angular distance $\left(\theta \right)$ covered by each planet is $2\pi$ . The linear velocity $\left(\nu \right)$ is evaluated as follows:

  $v=r\left(\frac{\theta }{t}\right)$

… (1)

Put the value of $r$ and $t$ of each planet in the equation (1) and solve their linear velocity.

  $\begin{array}{c}{v}_{Mercury}=\left(2440\right)\left(\frac{2\pi }{1407.6}\right)\\ =10.9km/h\\ \\ v_{Venus}=\left(6052\right)\left(\frac{2\pi }{5832.5}\right)\\ =6.5km/h\end{array}$

  $\begin{array}{c}{v}_{Earth}=\left(6356\right)\left(\frac{2\pi }{23.935}\right)\\ =1668.5km/h\\ \\ v_{Mars}=\left(3375\right)\left(\frac{2\pi }{24.623}\right)\\ =861.2km/h\end{array}$

Thus the linear velocity of Mercury, Venus, Earth and Mars is $10.9km/h,6.5km/h,$

  $1668.5km/h$ and $861.2km/h$ respectively.

b.

To determine

To compare: the linear velocity of a point on Mars equator with that of Earths equator.

Answer to Problem 40E

Linear velocity of Earth is almost double to that of Mars.

Explanation of Solution

Given information:

The time of rotation $\left(t\right)$ and radius $\left(r\right)$ of first four planets of solar system is shown below.

Planet	Radius (in kilometres)	Time for one rotation (hours)
Mercury	2440	1407.6
Venus	6052	5832.5
Earth	6356	23.935
Mars	3375	23.623

  $\begin{array}{c}{v}_{Earth}=1668.5km/h\\ v_{Mars}=861.2km/h\end{array}$

Calculation:

The linear velocity of earth and mars is compared below.

  $\begin{array}{c}\frac{v_{Earth}}{v_{Mars}}=\frac{1668.5km/h}{861.2km/h}\\ \frac{v_{Earth}}{v_{Mars}}=1.94\\ v_{Earth}\approx 2\left(v_{Mars}\right)\end{array}$

Thus the linear velocity of Earth is almost double to that of Mars.