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A body is lanuched at escape velocity from the Earth's surface i.e 11.2 km/s ,what would be it's velocity outside the Earth's gravitational field?

I guess it would be zero but there's another problem that is kind of confusing.

The problem is that this time a body is launched with thrice the escape velocity i.e. 33.6 km/s , so when it reaches at a point outside the Earth's gravitational field it should have a residual velocity of 33.6 - 11.2 = 22.4 km/s but the answer says it is 31.4, I don't get it.

Note - gravitational force due to sun and all the other planets is zero.

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    $\begingroup$ You do realize that energy, not speed, should be conserved? $\endgroup$ – my2cts Dec 6 '20 at 13:52
  • $\begingroup$ You need to look at the equations of motion to understand why speed is not conserved here. $\endgroup$ – Max Dec 6 '20 at 13:54
  • $\begingroup$ @my2cts I know energy is conserved, using energy conservation the book reached the answer 31.4 km/s but if my first assumption is true i.e residual velocity is zero when launched at escape velocity, then the answer should be 22.4 km/s $\endgroup$ – Fardeen Khan Dec 6 '20 at 14:08
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    $\begingroup$ "outside the Earth's gravitational field" What do you mean? The gravitational potential is inversely proportional to distance, it never goes to zero. Also, are you ignoring air resistance in this question? $\endgroup$ – PM 2Ring Dec 6 '20 at 14:09
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    $\begingroup$ So you know energy is conserved, yet your answer violates conservation. You should draw your conclusion. $\endgroup$ – my2cts Dec 6 '20 at 16:21
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The reason why the answer isn't $22.4$ km/s is because that doesn't satisfy the conservation of energy. You are not allowed to simply subtract the velocities. Your initial mechanical energy is $-GMm/r + mv^2/2$, and the final mechanical energy is $mv_f^2/2$, where $v_f$ if the final velocity. Energy isn't proportional to the velocity but to the velocity squared, so you can't just subtract the velocities. You can, however, subtract the squares of the velocities from each other and then take the square root to get the final velocity.

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  • $\begingroup$ Yeah, thank you, conserving velocity was a bad idea. $\endgroup$ – Fardeen Khan Dec 8 '20 at 16:37
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this wouldn't have been a problem if you had known how the formula is derived, instead of taking it as it is.

So basically, the gravitational potential energy of an object of mass M on the surface of the earth is -63m Million joules. Since the potential energy at infinity is 0, you need to increase the total energy of the object to 0 so that it can eventually escape the earth's field. This means that you need to give the object a k.e of +63M million joules. Using the kinetic energy formula, the M's cancel out, and the velocity found is 11.2 km/s.

Now, when you triple the velocity, you increase the kinetic energy 9 times. this means that you gave it a total k.e of 9x (63M million joules). At infinity, 63M million J will have been converted into gravitational potential energy, leaving the object with a kinetic energy of 8x(63M million Joules). You can find the velocity corresponding to this, which is about 32 km/s.

I tried to dumb this down as much as I could because the error in your understanding is pretty fundamental and can cause huge problems later on. You were directly subtracting the velocity when instead you were supposed to be subtracting kinetic energies. Essentially, you were calculating A-B instead of Sqrt( A^2-B^2), which I'm sure you are aware are completely different things

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  • $\begingroup$ Yes, thank you, I get it now, conserving velocity was dumb. $\endgroup$ – Fardeen Khan Dec 8 '20 at 16:37

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