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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)