91Ó°ÊÓ

1. The height of satellite above the earth surface is $640×{10}^3\mathrm{m}$.
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,$v=\frac{2\mathrm{Ï„}\mathrm{Ï„}{R}_{Total}}{T}$

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.

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

The equation of velocity is,

$$\begin{gathered} v=\frac{2\mathrm{τ}\mathrm{τ}{R}_{Total}}{T} \\ =\frac{2\left(3.14\right)\left(6.37×{10}^{6}m+640×{10}^{3}m\right)}{\left(98.0×6.0\right)s} \\ =7.49×{10}^{3}m/s \end{gathered}$$

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

We know the equation for centripetal acceleration.

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

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