A particle is fired vertically upward with a speed of $15\,\mathrm{km}/\mathrm s$. With what speed will it move in interstellar space? Assume only earth's gravitational field.

So this is a question from my physics book (I am in 12th grade) and I came across this problem from the chapter "Gravitation". So if we look simply, it's pretty clear that the left over speed of the particle should be the velocity of the particle moving freely in space which is $$\text{velocity of particle} -\text{escape velocity of earth}\\15\,\frac{\mathrm{km}}{\mathrm s}-11\,\frac{\mathrm{km}}{\mathrm s}=4\,\frac{\mathrm{km}}{\mathrm s}$$But the solution printed is $10\,\mathrm{km}/\mathrm s$ which looks absolutely unreasonable!


Velocity of the particle on the earth's surface $=v_e=15\,\mathrm{km}/\mathrm s$
Let $v_s$ be the particle's velocity in space.
Total energy must be conserved.
Change in $\text{K.E.}$ = change in $\text{P.E.}$
$$\Delta\text{K.E.}=\Delta\text{P.E.}\\\frac12m(v_e^2-v_s^2)=\frac{GMm}{R}\\\frac12(225-v_s^2)=\frac{6.67\cdot10^{-11}\cdot6\cdot10^{24}}{6400\cdot10^5}\\v_s=10\,\mathrm{km}/\mathrm s$$

Only $5\,\mathrm{km}/\mathrm s$ velocity used in escaping the Earth's gravitational field?!
This solution looks reasonable but not the answer.
Even the translunar injection performed by Saturn-V for moon was $10\,\mathrm{km}/\mathrm s$, the ISS's speed is $7.5\,\mathrm{km}/\mathrm s$, then how is it possible to send an object to space using just $5\,\mathrm{km}/\mathrm s$ speed?

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    $\begingroup$ Although you think it is pretty clear, the formula you suggested for finding the velocity in interstellar space is wrong. When leaving Earth, kinetic energy is converted into potential energy, and velocity enters the kinetic energy quadratically, as you can see in the solution and the answer by J.G.. So you cannot simply subtract one velocity from the other to find the resulting velocity in interstellar space. $\endgroup$
    – Koschi
    Commented Aug 1, 2021 at 13:17
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    $\begingroup$ By the way: The way you formulate the question title is probably confusing you. No one said that 5km/s is enough to escape. That is why the rocket starts with 15km/s. For example: If the rocket would start with 100km/s, the velocity in interstellar space would be roughly 99km/s, but that would also not mean that 1km/s is enough to escape. $\endgroup$
    – Koschi
    Commented Aug 1, 2021 at 13:22
  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Commented Aug 1, 2021 at 14:16
  • $\begingroup$ Although this looks like a "check my work" homework question, the OP isn't actually asking us to check the calculation. They want to know why the textbook uses that equation involving KE & PE rather than a simple difference of velocities. That is, the OP needed conceptual help on the connection between escape velocity, kinetic energy, and potential energy. $\endgroup$
    – PM 2Ring
    Commented Aug 1, 2021 at 18:15

2 Answers 2


So if we look simply the its pretty clear that the left over speed of the particle should be the velocity of the particle moving freely in space which is velocity of particle - escape velocity of earth

Why did you think you can find the required velocity by subtracting those velocities, simply like you ate one-sixth of a cake, then you have five-sixths left? Velocity is not like a cake :)

Objects don't gain or lose velocities due to work done on them. They gain or lose kinetic energy which has the term $v^2$ (velocity squared).

It seems you are in a misunderstanding that the velocity is being spared to escape from the earth's gravitational field. No, it's not. It's the kinetic energy that is being spared. So you can't get the answer by simply subtracting those velocities.


By energy conservation, escaping Earth's gravity subtracts a fixed amount of energy per mass (namely the GPE gained), i.e. of speed squared, not speed.

  • $\begingroup$ Oh my god my whole life was a lie, so 11 Km/s is just like a ticket to cross that gravitational sphere, all 11 Km/s isnt used to throw the body away, my brain hurts. Thanks $\endgroup$ Commented Aug 1, 2021 at 15:13

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