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Answer 1 · 2017-10-04T22:06:58

The orbital period is 3200 seconds (53.3 minutes).

We use Kepler's third law to solve this problem:

$T^2=((4pi^2)/(GM))r^3$
where G is the gravitational constant $6.67xx10^(-11)$

We find $M$ and $r$ from the information in the problem:

$M=18xx(1.90xx10^(27))= 3.42xx10^(28)$ kg

and $r = 1.2xx(6.99xx10^7) = 8.39xx10^7$ m

(Note that by "low orbit", we are considering an altitude just above the surface of the planet.)

Putting this all together, we get the equation:

$T^2=((4pi^2)/((6.67xx10^(-11))(3.42xx10^(28))))(8.39xx10^7)^3$

$T^2=(1.73xx10^(-17))(5.91xx10^(23))=1.02xx10^7s^2$

and finally

$T=3.20xx10^3$ s

(just over 53 minutes)