A particle is leaving the Moon in a direction that is radially outward from both the Moon and Earth.

What speed must it have to escape the Moon's gravitational influence?

Express your answer to two significant digits and include the appropriate units.

Explanation:

Essentially, Kinetic energy of the particle must equal the combined potential energies of earth and the moon when the object is on the moon's surface, meaning the full equation is

=Mass of Earth=

=Mass of Moon=

=distance from earth's center to the moon's=

=radius of moon=

After some algebra, the equation simplifies to

Plugging in the values of G, which is , should yield the proper answer of 2780m/s.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer