Question

During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of about 126km. How long did it take to go around the Moon once?

Answer 1

Answer:

It will take the command module 7159 seconds, or about 2 hours to go around the moon.

Explanation:

We use Kepler's third law which says the period $T$ of an orbit is

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

where $r$ is the distance from the center of the moon, $G$ is the gravitational constant, and $M$ is the mass of the moon.

Now,

$r = (1.73*10^6m)+(126*10^3m)$

$M = 7.35*10^{22}kg$

$G = 6.7*10^{-11} N m^2/kg$

putting these into Kepler's law, we get:

$T ^2 = \dfrac{4\pi^2 }{(6.7*10^{-11})(7.35*10^{22})} ((1.73*10^6m)+(126*10^3m))^3$

$T^2 = 51,254,442 s^2$

$\boxed{ T = 7159s}$

which is 119 minutes or 1 hour 59 minutes, which is approximately 2 hours.

Answer 2

The time taken by the command module to orbit the moon at an altitude of 126 km is 64 minutes 10 seconds

Time taken in the orbit:

The relation between the time period of a satellite in orbit and the radius of the orbit is given by:

$T^2=\frac{4\pi^2R^3}{GM}$

where R is the radius of the orbit

G is the gravitational constant = 6.7×10⁻¹¹Nm²/kg

M is the mass of the Moon = 7.35×10²² kg

Now, the radius of the orbit is:

R = radius of the moon + altitude of the satellite from the moon's surface

R = 1.73×10⁶ m + 0.126×10⁶ m

R = 1.856×10⁶ m

Substituting the respective values we get:

$T^2=\frac{4\pi^2(1.856\times10^6)^3}{6.7\times10^{-11}\times7.35\times10^{22}M}\\\\T^2= 14.86\times10^6s$

T = 3.85×10³s

T = 64 minutes and 10 seconds

Learn more about orbital motion:

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