gravitational potential energy between moon and earth.

Question

a) At what distance from the center of the earth along a line from the earth to the moon, is there no net force on a 10,000kg spacecraft? This is the crossover point at with the attraction to the moon becomes larger than the attraction to the earth.

b) The gravitational potential energy is the sum of the potential energies with respect to the earth and the moon. Calculate the potential energy of the spacecraft at several points between the surface of the earth and the surface of the moon. Then draw a reasonable graph showing the potential energy from the earth to the moon.

Answer 1

(a) The gravitational force between two bodies is F_g = G*m₁*m₂/R²

 

Part (a) says our spaceship is somewhere on a line between the earth and the moon, at a point where there is no net force on the ship.  It's always good to start out by drawing a free body diagram and identifying what forces are acting on our spaceship.  The only two forces here are the force of gravity due to the earth and the force of gravity due to the moon.  Because we are between the two, these forces will be acting in opposite directions.  Since we know the net force is zero, the magnitudes of these two gravitational forces must be equal.  So, how do we figure out the position?

 

 

Let's let r represent the distance from the earth to the space ship.  The force of gravity due to the earth will be

F_e = G*m_e*m_s/r²

Here m_e is the mass of the earth, m_s is the mass of our ship, and G is the gravitational constant.

 

Now, we need the force of gravity due to the moon.  Be careful with the denominator, this is the distance between the two bodies.  And r is the distance from the ship to the earth, not the distance from the ship to the moon.  So it's going to be (r_m - r) where r_m is the distance from the earth to the moon.

F_m=G*m_o*m_s/(r_m - r)²

Here m_o is the mass of the moon (I picked m_o because m_m is a little confusing)

 

The magnitude of these two forces are equal so

 

Fe = Fm

G*m_e*m_s/r² = G*m_o*m_s/(r_m - r)²

 

There's G and m_s on both side of the equation, so they cancel out

 

m_e/r² = m_o/(r_m - r)²

 

Now let's get all of the mass terms on one side of the equation and all the distance terms on the other.  To do this, multiply both sides by (r_m - r)² and divide both sides be m_e

 

(r_m - r)²/r² = m_o/m_e

 

Take the square root of both sides

 

√((r_m - r)²/r²) = √(m_o/m_e)

(r_m - r)/r = √(m_o/m_e)

r_m/r - r/r = √(m_o/m_e)

r_m/r - 1 = √(m_o/m_e)

 

Move the one over to the other side, then invert to get r in the numerator

rm/r = √(mo/me) + 1

r/r_m = 1/(√(m_o/m_e) + 1)

 

Lastly multiply both sides by rm (the distance from the earth to the moon)

 

r = r_m/(√(m_o/m_e) + 1)

 

 

(b) After the last one, this part is pretty straightforward.  The gravitational potential energy is U = -G*m₁*m₂/R

 

As before, we need to make sure we using the correct R.  So, let r be the distance from the earth to the spaceship.

 

U_e = -G*m_e*m_s/r

 

If r is the distance from the earth to the spaceship, then the distance from the spaceship to the moon is (r_m - r) where r_m is the distance from the earth to the moon (just like before).  That means the gravitational potential energy due to the moon is

 

U_o = -G*m_o*m_s/(r_m - r)

 

As is says in the question, the total potential energy is just the sum of those two

U = U_e + U_o

 

All you have to do now is pick some points and plug in the numbers.  Quick tip, make sure one of the r values you use is the r from part (a).  Also, keep in mind that as you approach either the earth or the moon, you're getting farther away from the other one.  So while U_e goes down (because you're getting closer to the earth), U_o will go up (because you're getting farther from the moon) and vice versa.