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Work and Potential Energy: To the Moon physics problem
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[QUOTE="Zeus5966, post: 4541060, member: 491325"] [h2]Homework Statement [/h2] Part 1(complete and correct. I'm including this for context's sake) You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the Earth with enough speed to make it to the moon. You leave the surface of the Earth at v = 5534m/s. How far from the center of the Earth will you get? [B]8447116.43[/B] 2)Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin=11068m/s. How fast will you be moving at the surface of the moon? Variables: Vmoon = ? Vmin = [B]11068m/s[/B] 3) Which of the following would change the minimum velocity needed to make it to the moon? (I'm pretty sure this is correct, however let me know if I'm wrong) Mass of the Earth *True* Radius of the Earth *True* Mass of the spaceship *False* [h2]Homework Equations[/h2] Ue = -(G*Me*m)/rem where rem is the radius of the Earth plus the distance between Earth and moon Kf = (1/2)mVf^2 Ki = (1/2)mVi^2 Um = -(G*Mm*m)/rm where rm is the radius of the moon Where Me is the mass of earth, Mm is the mass of the moon, m is the mass of the spaceship(which does not matter as it will be canceled in all terms) and Vi is the minimum escape velocity needed. [h2]The Attempt at a Solution[/h2] Part 2 Ue + Ki = Um + Kf Ue - Um + Ki = Kf -G*Me/(rem) - -G*Me/(rm) + V^2/2 = Vf^2/2 √2*(-G*Me/(rem) - -G*Me/(rm) + V^2/2) = Vf Calculating this gives me 21365.16m/s which is an incorrect answer. Thank you [/QUOTE]
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Work and Potential Energy: To the Moon physics problem
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