At some point along the direct path from the center of the earth to the center of the moon, the gravitational force of attraction on a spacecraft from the moon becomes greater than the force from the earth. a) how far from the center of the earth does this occur? b) at this location, how far is the spacecraft from the surface of the moon? c) how far is it from the surface of the earth?

a)  r₁ = 3.4565 10⁸ m , b)    = 3.661 10⁷ m , c)    = 3.39 10⁸ m

Explanation:

a) For this exercise we will use Newton's second law where the forces are of gravitational attraction

         F = G m M / r²

Let's locate the point where the forces are equal and from here we know which way they are greater

        F₁ - F₂ = 0

F₁ is the force of attraction from Earth and F₂ the force of attraction from the Moon, let's use the subscript "e" for the Earth and the subscript "m" for the Moon

       G m / r₁² = G m / r₂²

        / = (r₁ / r₂)²

Let's measure the distance from the same reference system, which we will place in the center of the Earth

     r₁ = r₁

    r₂ = D - r₁

Where D is the distance from beats to the Moon, substitute

     r₁ / D-r₁ = √ /

     r₁ / D-r₁ = √ (5.98 1024 / 7.36 1022) = 9.0139

     r₁ = 9.0139 (D -r₁)

    r₁ (1+ 9.0139) = 9.0139 D

    r₁ = 9.0139 / 10.0139 D

    r₁ = 0.9001 3.84 10⁸

    r₁ = 3.4565 10⁸ m

It is this point sides forces are equal for greater distances the force of the moon exceeds the attraction of the Earth

b) The radius of the moon is

      = 1.74 10⁶ m

The distance measured from the Moon is

      r₂ = D -r₁

      r₂ = 3.84 108 - 3.4565 108

      r₂ = 0.3835 108 m

      r₂ = 3.835 107 m

The distance r₂ is measured from the center of the Moon, the distance measured from the surface is

       = r₂ -  

       = 3.835 10⁷ - 1.74 10⁶

       = 3.661 10⁷ m

c)   The distance from the Earth's surface is

      = r1 -  

      = 3.4565 10⁸ - 6.37 10⁶

     = 3.39 10⁸ m

answer; i believe the correct answer is (40,000 j) good luck

0.55x745.7j and a calculator