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ANTONII [103]
3 years ago
9

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

Physics
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

2780m/s

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

<h3>\frac{1}{2} mv^2=\frac{G(M_E)m}{r_E} +G\frac{M_mm}{r_m}\\</h3><h3 />

M_E=Mass of Earth=5.97*10^2^4

M_m=Mass of Moon=7.4*10^2^2kg

r_E=distance from earth's center to the moon's=3.84*10^8m

r_m=radius of moon=1.738*10^6m

After some algebra, the equation simplifies to

v=\sqrt{2G*(\frac{M_E}{r_E+r_m}+\frac{M_m}{r_m})}

Plugging in the values of G, which is 6.67*10^-^1^1 \frac{m^3}{kg*s^2}, should yield the proper answer of 2780m/s.

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