Question

A satellite is in a circular orbit around the Earth at an altitude of 2.80 3 106 m. Find (a) the period of the orbit, (b) the speed of the satellite, and (c) the acceleration of the satellite. Hint: Modify Equation 7.23 so it is suitable for objects orbiting the Earth rather than the Sun.

Answer 1

Explanation:

Given that,

Altitude $h= 2.803\times10^{6}\ m$

We need to calculate the radius

$r=R+h$

Where, R = radius of the earth

h = radius of altitude

Put the value into the formula

$r=(6.38\times10^{6}+2.803\times10^{6})$

$r=9.18\times10^{6}\ m$

(a). We need to calculate the period of the orbit,

Using formula of period

$T^2=\dfrac{4\pi^2}{GM}r^3$

$T^2=\dfrac{4\pi^2}{6.67\times10^{-11}\times5.98\times10^{24}}\times(9.18\times10^{6})^3$

$T^2=76570372.9509\ sec$

$T=8750.45\ sec$

(b). We need to calculate the speed of the satellite

Using formula of speed

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

Put the value into the formula

$v^2=\dfrac{6.67\times10^{-11}\times5.98\times10^{24}}{9.18\times10^{6}}$

$v^2=43449455.3377\ m/s$

$v=6.59\times10^{3}\ m/s$

(c). We need to calculate the acceleration of the satellite

Using formula of acceleration

$a_{c}=\dfrac{v^2}{r}$

Put the value into the formula

$a_{c}=\dfrac{(6.59\times10^{3})^2}{9.18\times10^{6}}$

$a_{c}=4.73\ m/s^2$

Hence, This is the required solution.