Question

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and 270 km above the surface of the Moon, where the acceleration due to gravity is 1.27 m/s2. The radius of the Moon is 1.70 ✕ 106 m. Determine the astronaut's orbital speed.

Answer 1

Answer:

$v  =  1,582 \ \frac{m}{s}$

Explanation:

We know that for circular motion the centripetal acceleration $a_c$ is:

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

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

$a_c = 1.27 \frac{m}{s^2}$.

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

$r = 1.7 * 10^6 \ m + 270  \ km$

$r = 1.7 * 10^6 \ m + 270 * 10^3 \ m$

$r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m$

$r = 1.97 * 10^6 \ m$

We can obtain the speed as:

$v^2  = a_c r$

$v  = \sqrt{a_c r}$

$v  = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}$

$v  = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}$

$v  =  1.582 \ 10^3 \ \frac{m}{s}$

$v  =  1,582 \ \frac{m}{s}$

And this is the orbital speed.