Asked by James
Standing on the surface of a small spherical moon whose radius is 6.30 104 m and whose mass is 8.00 1018 kg, an astronaut throws a rock of mass 2.02 kg straight upward with an initial speed 42.2 m/s. (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?
The answer is 7400 m. Could someone show me the steps to get this answer? Thank you in advance.
The answer is 7400 m. Could someone show me the steps to get this answer? Thank you in advance.
Answers
Answered by
Steve
Gravtational acceleration is = GM/r^2
At the surface , that is
6.673*10^-11 * 8.00*10^18 / (6.30*10^4)^2 = 0.1345
h = Vot - 1/2 at^2
= 42.2t - .06725 t^2
h=0 at t=0, 627.5
h is max halfway between the roots, at t = 313.75
h(313.75) = 6620.223
Check my typing and math -- I don't get 7400.
At the surface , that is
6.673*10^-11 * 8.00*10^18 / (6.30*10^4)^2 = 0.1345
h = Vot - 1/2 at^2
= 42.2t - .06725 t^2
h=0 at t=0, 627.5
h is max halfway between the roots, at t = 313.75
h(313.75) = 6620.223
Check my typing and math -- I don't get 7400.
Answered by
James
I'm getting a negative answer when I checked your math. Also, shouldn't the mass of the rock be considered?
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