Question #a4f57

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Question #a4f57

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Answer 1 · 2017-10-04T22:06:58

The orbital period is 3200 seconds (53.3 minutes).

Explanation:

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

$T^{2} = (\frac{4 π^{2}}{G M}) r^{3}$ $T^2=((4pi^2)/(GM))r^3$
where G is the gravitational constant $6.67 \times 10^{- 11}$ $6.67xx10^(-11)$

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

$M = 18 \times (1.90 \times 10^{27}) = 3.42 \times 10^{28}$ $M=18xx(1.90xx10^(27))= 3.42xx10^(28)$ kg

and $r = 1.2 \times (6.99 \times 10^{7}) = 8.39 \times 10^{7}$ $r = 1.2xx(6.99xx10^7) = 8.39xx10^7$ 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:

$T^{2} = (\frac{4 π^{2}}{(6.67 \times 10^{- 11}) (3.42 \times 10^{28})}) {(8.39 \times 10^{7})}^{3}$ $T^2=((4pi^2)/((6.67xx10^(-11))(3.42xx10^(28))))(8.39xx10^7)^3$

$T^{2} = (1.73 \times 10^{- 17}) (5.91 \times 10^{23}) = 1.02 \times 10^{7} s^{2}$ $T^2=(1.73xx10^(-17))(5.91xx10^(23))=1.02xx10^7s^2$

and finally

$T = 3.20 \times 10^{3}$ $T=3.20xx10^3$ s

(just over 53 minutes)

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Question #a4f57

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