anonymous
  • anonymous
A hypothetical planet has a mass 1.90 times that of Earth, but the same radius. What is g near its surface?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Use that the force is \[F=G*m*m p/r ^{2}\] with G=6.67e-11, m the mass of your point and mp the mass of the planet. Then consider you want your force to be some \[F=m*g\]. So : \[g = G*m p/r ^{2}\]
anonymous
  • anonymous
When I plug in my values I get: \[6.673*10^-11Nm^2/kg^2*1.136kg/(6.38*10^3km)^2=1.86m/s^2\] I must have gotten the wrong mass of the planet because its saying my answer is wrong. I did this to get the mass of the planet: \[5.98*10^{24}*1.90=1.136kg\]
anonymous
  • anonymous
Is this the answer: \[g=Gm _{p}/r^2(6.61*10^-11Nm^2/kg^2)(5.98*10^24kg)(1.1362*10^25)/(6.38*10^3km)^2\]=\[18.62g\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
nevermind the answer was 18.6m/s

Looking for something else?

Not the answer you are looking for? Search for more explanations.