To measure the escape velocity, if I use the equation -

$$\frac{-GMm}{r^2}=\frac{v.dv}{dr} $$

and I put my final distance to be $\infty$ , then I get the answer,

$$u = \sqrt\frac{2GM}{R} \;. \tag{1}$$ Which is quite obvious!

But, if I use the equations -

$$U_{\infty} - U_i = \frac{GMm}{R}$$


$$K_{\infty}+U_{\infty} = K_R+U_R$$

Where, $K_R$ and $U_R$ are kinetic energies and potential energies at R respectively , where R is the radius of the earth.

Now in these two equations, if I put $U_{\infty} = 0$, then I get,

$$K_{\infty} = K_R - \frac{GMm}{R}\tag{3} $$

And since, $K_{\infty}$ is positive and not zero, we end up getting that

$$u > \sqrt\frac{2GM}{R} \tag{2} $$

That is to say that, if I want to take the object at infinity, the speed must be (1), but if I want to make it's final potential energy 0 , then it's speed must be greater than (1). Also, (1) cannot be used in (3) because that would mean the final $KE$ is 0, which obviously is not!

What is generating this hoax? Please tell me where am I wrong?

  • $\begingroup$ I re-worked a lot of your LaTeX. Please check that I didn't mess anything up, but also look at what I did. You can use $$ delimiting to block-set equation, put the whole equation in a single set of delimiters, and use \tag to set equation numbers. $\endgroup$ – dmckee --- ex-moderator kitten Dec 2 '16 at 20:19
  • $\begingroup$ @dmckee thank you! Actually, I do not know how to type it effectively ! I hope that next time, I will do it better. $\endgroup$ – Aaryan Dewan Dec 3 '16 at 3:07

The escape velocity is the minimum velocity possible such that it escapes a gravitational well. We have from conservation of energy: $$ E = K+U = \frac{1}{2}mv^2 - \frac{GMm}{r} $$

Lets consider initial and final energy as $E_1$ and $E_2$. To escape, you need to make you sure your speed fully overrules the potential energy. The maximum possible potential energy there is, is the case $U = 0$, that happens a $r\to\infty$. At all other cases, we have $U < 0$, and thus $U = 0$ is maximum. If your kinetic energy is greater than this maximum, you escaped. So, lets make sure of this. Assuming your initial energy to be: $$ E_1 = K_1 + U_1 = K_2 + U_2 = E_2 $$

At maximum, we have $U_2 = 0$, then: $$ K_1 + U_1 = K_2 \quad\implies\quad K_1 = K_2 - U_1 $$

So, if you have kinetic energy $K_1$, you escape. As you pointed out, the things at the end happens as $r\to\infty$ and therefore $E_2 = E_\infty = K_\infty + U_\infty$. Thus: $K_2 = K_\infty$. That is, that will be your kinetic energy at infinity. $$ K_1 = K_\infty - U_1\quad\implies\quad \frac{1}{2}mv^2 = \frac{1}{2} mv_\infty^2 + \frac{GMm}{r_1} $$

The velocity you need to escape such that, at infinity, you have velocity $v_\infty$ is: $$ v = \sqrt{v_\infty^2 + \frac{2GM}{r_1}} $$

At $v_\infty = 0$, you recover the minimum velocity you need to have in order to escape. Since it is minimum, your velocity at infinity will be zero. Once you arrive at infinity, regardless if you have velocity or not, you escape, and then your potential energy is zero.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the answer! So What you're saying is that the final KE and PE is 0 and as we know that the initial PE is negative, so our initial KE must somehow balance our initial Potential energy, so the total energy of our rocket at any instant must be zero! ? $\endgroup$ – Aaryan Dewan Dec 2 '16 at 19:15
  • 1
    $\begingroup$ I am saying that, if you have any energy $E\ge 0$, you escape. In particular, if $E = 0$, then you escape, and at infinity your kinetic energy (and thus velocity) is zero. So, the escape velocity is precisely when $E = 0$, since escape velocity is the minimum velocity needed to escape. Any velocity greater will do as well. =). $\endgroup$ – Physicist137 Dec 2 '16 at 19:19

We define escape velocity to be the minimum velocity required to just overcome the earth's pull. So it is assumed that at infinity the kinetic energy is 0. Thus both the equations give correct answers. Although we are neglecting the air drag and other forces. So in practical cases the initial velocity , u >(2GM /R)^(1/2). But theoritically escape velocity is (2GM/R)^(1/2)... from both the equations.

| cite | improve this answer | |

So, it all depends upon the initial energy of the system. If the final KE and PE are 0 just as the accepted answer tells you, then you need to use the precise value of escape velocity so that it exactly cancels out with $\frac{-GMm}{R}$. Then according to this at infinity, you will come to rest.

But if you get some more initial energy by increasing your initial KE, then you will reach infinity, your final potential energy will become 0 but you will never come at rest.

So there is no mathematical inconsistency in the question. I was just not able to understand what results am I getting.

| cite | improve this answer | |
  • $\begingroup$ You are still confused. Escape velocity means the minimum velocity to get to infinity. Setting down the right conditions to interpret how to solve a physics problem is part of understanding physics or maybe logic or words. $\endgroup$ – Bob Bee Dec 3 '16 at 4:41
  • $\begingroup$ Can you please point out, where am I wrong in the answer that i provided? $\endgroup$ – Aaryan Dewan Dec 3 '16 at 4:42
  • $\begingroup$ Escape velocity means zero velocity at infinity. You answer about mathematical inconsistency makes no difference. Just accept that the right answer, as you seem to have chosen, is right. No if and buts. $\endgroup$ – Bob Bee Dec 3 '16 at 4:45
  • $\begingroup$ Thanks @BobBee ! But can you please tell me that if i push my object with an initial velocity greater than the escape velocity, then it's final velocity at infinity will not be zero, as I pointed out? $\endgroup$ – Aaryan Dewan Dec 3 '16 at 4:51
  • $\begingroup$ Yes, clearly. Yes. $\endgroup$ – Bob Bee Dec 3 '16 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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