Kepler's laws and orbits question?

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SUMMARY

The discussion centers on calculating the gravitational field strength at the surface of an imaginary planet with a diameter twice that of Earth. Using Kepler's third law, the gravitational parameter GM was computed as 6.61 x 10^21 m^3/s^2. Subsequently, the gravitational field strength (g) was determined to be approximately 40,597,576.67 N/kg, which is significantly higher than Earth's gravity, leading to concerns about the feasibility of life on such a planet.

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Homework Statement



An imaginary planet has a diameter twice that of our earth. The planet has one moon with an orbital radius of 5.0x10^7 km and orbital period of 10 Earth days.
Determine the strength of the gravitational field at the surface of the planet.

Radius of Earth = 6.38 x 10^6m
Orbital radius of moon = 5x10^10m
orbital period = 8.64 x 10^5 s

Homework Equations



r^3/T^2 = GM/4pi^2
g = GM/r^2

i think that's all..?...

The Attempt at a Solution



GM = (4pi^2(5x10^10)^3)/(8.64x10^5)^2
= 6.61 x 10^21

g = 6.61x10^21/(12.76x10^6)^2
= 40597576.67N/kg

this number is massive! I must be doing something wrong. any help?
 
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Your calculation looks fine. I wouldn't want to live there!
 

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