An Earth satellite moves in a circular orbit 640 km ( uniform circular motion ) above Earth’s surface with a period of 98.0 min . What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

1. The height of satellite above the earth surface is $640\times {10}^3\mathrm{m}$.
2. Time of that satellite is 98.0 min.
3. As well as he radius of earth is$6.37x{10}^{6}m$

As given in the problem, the earth satellite is moving with a circular orbit around the earth at a certain height. So, by using the radius of the earth and given parameters, we can easily find the desired values.

Formulae:

Speed of satellite,$v=\frac{2\tau \tau R_{Total}}{T}$

Acceleration,

$a=\frac{v^2}{r}$

Let’s take the radius of earth as the total radius of system. So, we can find the velocity of satellite. Once we get the value of velocity, it is easy to calculate magnitude of acceleration.

$R_{Total}=6.37\times {10}^{6}m+640\times {10}^{3}m$

The equation of velocity is,

$$\begin{gathered} v=\frac{2\tau \tau R_{Total}}{T} \\ =\frac{2\left(3.14\right)\left(6.37\times {10}^{6}m+640\times {10}^{3}m\right)}{\left(98.0\times 6.0\right)s} \\ =7.49\times {10}^{3}m/s \end{gathered}$$

Therefore, the speed of the satellite is $7.49x{10}^{3}m/s$.

We know the equation for centripetal acceleration.

$$\begin{gathered} a=\frac{v^2}{R_{Total}} \\ =\frac{{\left(7.49\times {10}^{3}m/s\right)}^2}{\left(6.37\times {10}^{6}m+640\times {10}^{3}m\right)} \\ =8.00m/s^2 \end{gathered}$$

Therefore, the magnitude of the centripetal acceleration is $8.00m/s^2$.