Answer to Question #198585 in Mechanics | Relativity for Mthoko

Question

A satellite orbits the earth a distance of 1.50 × 10⁷ m above the planet's surface and takes 8.65 hours for each revolution about the earth. The earth's radius is 6.38 × 10⁶ m. The acceleration of this satellite is closest to

Answer 1

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r} \;(1)$

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}\; (2)$

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite$(1.50 \times 10^{7} \;m)$ and the Earth radius $(6.38 \times 10^{6} \;m)$ :

$r = 1.50 \times 10^{7} \;m+6.38 \times 10^{6} \;m \\ r = 21.38 \times 10^{6} \;m$

Then, equation 2 can be used:

$T = 8.65 \;hrs \times \frac{3600 \;s}{1\;hrs}=31140 \; s \\ v = \frac{2 \pi (21.38 \times 10^{6}\;m)}{31140 \;s} \\ v = 4313 \;m/s$

Finally equation 1 can be used:

$a = \frac{(4313 \;m/s)^{2}}{21.38 \times 10^{6}m} \\ a = 0.87\; m/s^{2}$

Hence, the acceleration of the satellite is $0.87 \;m/s^{2}$