Part one

on the way to the moon, the apollo astronauts reach a point where the moon’s gravitational pull is stronger than that of earth’s.
find the distance of this point from the
center of the earth. the masses of the
earth and the moon are 5.98 × 10^24 kg and
7.36 × 10^22 kg, respectively, and the distance
from the earth to the moon is 3.84 × 10^8 m.
answer in units of m.

part two

what would the acceleration of the astronaut be due to the earth’s gravity at this
point if the moon was not there? the
value of the universal gravitational constant
is 6.672 × 10^−11 n · m^2/kg^2
answer in units of m/s^2

3.64×10⁸ m

3.34×10⁻³ m/s²

Explanation:

Let's define some variables:

M₁ = mass of the Earth

r₁ = r = distance from the Earth's center

M₂ = mass of the moon

r₂ = d − r = distance from the moon's center

d = distance between the Earth and the moon

When the gravitational fields become equal:

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

M₁ / r² = M₂ / (d − r)²

M₁ / r² = M₂ / (d² − 2dr + r²)

M₁ (d² − 2dr + r²) = M₂ r²

M₁d² − 2dM₁ r + M₁ r² = M₂ r²

M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0

d² − 2d r + (1 − M₂/M₁) r² = 0

Solving with quadratic formula:

r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)

When we plug in the values, we get:

r = 3.64×10⁸ m

If the moon wasn't there, the acceleration due to Earth's gravity would be:

g = GM / r²

g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²

g = 3.34×10⁻³ m/s²

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray