Free-Fall WITH Variational Gravity

[DThielke]

Problem background:

• Body free-falling towards the earth.
• Account for variation in gravitational force with height.
• Neglect air resistance.
• Initial conditions summarized by total energy E.

I'm having trouble understanding part of my textbook's solution to this problem, so I will paraphrase said solution up until I have trouble. My comments are in bold.

Beginning with

$$F=-\frac{mMG}{x^2}$$,​

it can be shown that

$$V(x)=-\frac{mMG}{x}$$.​

By application of conservation of energy, we determine

$$v=\frac{dx}{dt}=\sqrt{\frac{2}{m}}\left[E-V(x)\right]^\frac{1}{2}=\sqrt{\frac{2}{m}}\left[E+\frac{mMG}{x}\right]^\frac{1}{2}$$.​

In order to find x(t), we must evaluate

$$\int^x_{x_0}\frac{dx}{\left(E+\frac{mMG}{x}\right)^\frac{1}{2}}=\sqrt{\frac{2}{m}}t$$.​

I can easily follow the math up to this point, but the next step in the textbook solution throws me off.

To solve the case when E is negative, substitute

$$\cos{\theta}=\sqrt{\frac{-Ex}{mMG}}$$,​

and we arrive at

$$\frac{mMG}{\left(-E\right)^\frac{3}{2}}\int^\theta_{\theta_0}2\cos^2\theta d\theta=\sqrt{\frac{2}{m}}t$$.​

If I assume this substitution to be valid, I can easily follow the rest of the textbook to the solution. However, where does this seemingly random substitution come from? And can anyone show how it is mathematically applied to the integral? I don't feel right blindly accepting the trig substitution and would appreciate clarification.

Thanks

 

[ApexOfDE]

Because you say E is negative ==> -E > 0.

Lets consider:

$$\sqrt{E + \frac{mMG}{x}} = \sqrt{(-E)(-1 + \frac{mMG}{-Ex}$$

From this point, you should imagine that you need substitution to solve this integral. Usually, the best way to solve du/(1-u^2) is using trigonometry. You can try to put cos/sin(u) = sqrt(mMG\-Ex) and then using a substitution from your textbook you will find that which way is more better.

The more practice you do with this type of integral, the more sense you have when you find a best way, that's my exp.