Question

r²=a²-2Racosθ+R²
Verify this last expression, plug it into the following equation:
dV=(-GMmsinθdθ)/(2r)
and integrate from θ=0 to θ=π. (they suggest a change of variables: define u=cos θ) And we should find that the total value of V for the interaction is V=-GMm/R

Answer 1

The only thing that I'm not sure about is what it means by "verify this last expression."

From observation, it at least looks like it comes from law of cosines with sides a, r and R and angle theta. I'm not sure how else that would be verified unless plugging in values and checking.

Answer 2

It wants us to plug the first equation into the second equation and integrate it, hopefully coming up with an end equation of V=-GMm/R

Answer 3

After verifying the equation, you could certainly plug it in...
If we take the square root:
\( r = \sqrt{a^2 - 2Ra \cos \theta + R^2} \) r > 0

\( dV = \dfrac{-GMm \sin \theta \ d \theta}{\sqrt{a^2 - 2Ra \cos \theta + R^2}} \)
and integrate:
\( \displaystyle \int dV = \int_{0}^{\pi} \dfrac{-GMm \sin \theta \ d \theta}{\sqrt{a^2 - 2Ra \cos \theta + R^2}} \)
If we substitute \( u = \cos \theta \), we have \( du = -\sin \theta \ d \theta \), and:
\( \displaystyle V = \int_{1}^{-1} \frac{GMm \ du}{\sqrt{a^2 + R^2 - 2Ra u}} \)

Have you made it that far?

Answer 4

We could make one more substitution that would really simplify this:
\( t = a^2 + R^2 - 2R a u \)
\( dt = -2Ra \ du \to -\dfrac{1}{2R} dt = du\)

\( \displaystyle V = -\dfrac{GMm}{2R} \int_{t(1)}^{t(-1)} \dfrac{1}{\sqrt{t}} \ dt \)
\(\sqrt{t} = t^{1/2} \)
\( \displaystyle V = - \dfrac{GMm}{2R} \int_{t(1)}^{t(-1)} t^{-1/2} \ dt \)

Answer 5

Oops, I should be \( - \dfrac{1}{2Ra} \ dt = du \) and subsequently for all the fractions of 2R = 2Ra.