91Ó°ÊÓ

The given data can be listed below,

• The earth’s satellite is moving in a circular orbit.
• The orbit speed of earth’s satellite is$6200\mathrm{m}/\mathrm{s}$

Motion around a fixed point or body is orbital motion. The object's velocity has to be less than the speed needed to escape the "gravitational effect" of the planet for orbital motion.

Newton’s second law and law of gravitation force can be given by,

$G\frac{M\mathrm{m}}{r^2}=m\frac{v^2}{r}$

Here, Mis the mass of the planet whose value is $5.97×{10}^{24}\mathrm{kg}$,G is the constant of gravitation whose value is$G=6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$,m is the mass of the satellite, r is the radius of earth and v is the orbital speed of the satellite.

Solve the above equation for r.

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

Substitute values in the above,

$$\begin{gathered} \mathrm{r}=\frac{\left(6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.97×{10}^{24}\mathrm{kg}\right)}{\left(6200\mathrm{m}/\mathrm{s}\right)}\left(\frac{\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right) \\ =1.04×{10}^7\mathrm{m} \end{gathered}$$

The time taken for circular motion is given by,

localid="1657180469712" $T=\frac{2\mathrm{Ï€°ù}}{v}$

Substitute all values in the above,

$$\begin{gathered} \mathrm{T}=\frac{2\mathrm{π}\left(1.04×{10}^7\mathrm{m}\right)}{6200\mathrm{m}/\mathrm{s}} \\ =1.05×{10}^4\mathrm{s}\left(\frac{1\mathrm{hr}}{3600\mathrm{s}}\right) \\ =2.92\mathrm{hr} \end{gathered}$$

Thus, the time taken by the satellite to complete one revolution is$2.92\mathrm{hr}$.

The radial acceleration of the satellite can be expressed as,

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

Here, v is the orbital speed of the satellite, and r is the radius of the earth.

Substitute the values in the above,

$$\begin{gathered} a_r=\frac{{\left(6200m/s\right)}^2}{1.04×{10}^{7}m} \\ =3.70m/s^2 \end{gathered}$$

Thus, the radial acceleration of the earth’s satellite is $3.70m/s^2$.