Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. if you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon!

2.456 km /s

Explanation:

Let the mass of the object thrown be unity.

Potential energy due to attraction of the earth on the earth

=  - GM/R

(6.6 x 10⁻¹¹ x 5.972 x 10²⁴) / 6371 x 10³

= - 6.18 x 10⁷ J

Kinetic energy due to velocity of the object

= .5 x ( 11068 )²

= 6.1 x 10⁷ J

Total energy

= - .08 x 10⁷ J

Potential energy due to attraction of the moon on the moon

=  - GM/R

(6.6 x 10⁻¹¹ x 7.347 x 10²²) / 1737 x 10³

= - .279 x 10⁷ J

Potential energy due to attraction of the earth  on the moon

=  - GM/d where d is distance from the earth to the moon

=  (6.6 x 10⁻¹¹ x 5.972 x 10²⁴) / 3.84 x 10⁸

= -.1026 x 10⁷ J

Total potential energy

= -.3816 x 10⁷ J

If v be the velocity on the moon

total energy on the moon

= -.3816 x 10⁷  + .5 v² = -  .08 x 10⁷

.5 x v² = ( .3816 - .08 ) x 10⁷

v² = 6.032 x 10⁶

v = 2.456 x 10³m /s

= 2.456 km /s