0
$\begingroup$

I have tried integrating $v dv = \frac{-G M}{x^{2}} dx$ to get the popular $v = \sqrt{2 G M \left( \frac{1}{x} - \frac{1}{h} \right)}$.

After this, I used trigonometric substitutions to get time in terms of height. Since the equation obtained (written below) is transcendental, I can't express height as a function of time.

How do I express height as a function of time? Is there some other method of solving the differential equation or is the equation necessary entirely different?

The equation I am getting is $\sqrt{ \frac{h^3}{2 G M}} \left( \frac{\pi}{2} - \arcsin \left( \sqrt{\frac{x}{h} } \right) + \frac{1}{h} \sqrt{x \left( x - h \right) } \right) = t$

Improve this question
$\endgroup$
5
  • $\begingroup$ You don’t have to integrate anything to get your expression for $v$. It follows immediately from the conservation of energy. $\endgroup$
    – Ghoster
    Commented Jun 16, 2023 at 19:58
  • $\begingroup$ You can’t get an analytic expression for $x(t)$… only for $t(x)$. $\endgroup$
    – Ghoster
    Commented Jun 16, 2023 at 20:04
  • $\begingroup$ This question would be much clearer if you showed $t(x)$. Without it, readers won’t understand what trascendental equation you are talking about. $\endgroup$
    – Ghoster
    Commented Jun 16, 2023 at 20:06
  • $\begingroup$ @Ghoster I have added the equation showing t(x). Could you suggest some new method or new equation? Thanks $\endgroup$
    – Aashv
    Commented Jun 17, 2023 at 19:35
  • $\begingroup$ You can’t get a nice formula for $x(t)$. You can get an ugly series. Why is $t(x)$ not good enough? $\endgroup$
    – Ghoster
    Commented Jun 17, 2023 at 20:25

1 Answer 1

Reset to default
0
$\begingroup$

Based on my understanding of your question, I propose the following solution:

Applying Newton's Law: $$m\frac{d^2h}{dt^2}=-\frac{GMm}{(R+h)^2}$$ Where $R$=Radious of the earth

Solving the above differential equation, you get,

1

Follow this link for a detailed solution of the above differential equation

Where $C_1$ and $C_2$ can be obtained by applying 2 boundary conditions such as the following (it may vary as the case may be):

  1. At $t=0, h=0$
  2. At $t=0$, initial speed is known i.e. $\frac{dh}{dt}=u$
Improve this answer
$\endgroup$
2
  • $\begingroup$ I tried to view the detailed solution but couldn't do it. How was the differential equation solved? I tried to obtain a similar equation but couldn't do it. Thanks $\endgroup$
    – Aashv
    Commented Jun 17, 2023 at 19:57
  • $\begingroup$ You need to have pro account in order to view full steps. $\endgroup$
    – Vikash Kumar
    Commented Jun 19, 2023 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.