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When an object leaves a celestial body (e.g. planet or star) with it's escape velocity, according to first law of motion, it will continue travel away from the body with the same speed (without having need of any additional energy).

Let's assume that: At time t1 the object has at distance d1 from the celestial body. And At time t2 the object has at distance d2 from the celestial body.

Here t2 > t1 and d2 > d1

Now if the object falls from distance d1 back to the body, at the time of impact it will have certain energy.

Let's assume this energy is j1 Jules.

And if the object falls from distance d2 back to the body, at the time of impact it will have more energy than j1 (Because, d2 > d1 hence gravitational acceleration will cause more speed at the time of impact)

Let's assume this energy is j2 Jules

Question is:

This total system didn't consume any external energy (neither the object was supplied with additional fuel after it starts its journey from celestial body). Then how does it generate more energy just by travelling for more time in space? How can this be explained in the context of law of conservation of energy?

Update 1:

This is not question about escape velocity. It's more about how to explain relation of energy of impact of a trajectory (if it falls back) with the amount of time it has travelled in space. More the amount it travels, more is it's impact. How to explain that especially wrt Energy conservation law.

Update 2:

I was using some online free fall calculators (for e.g. this, this and this) and they calculated speed exceeding escape velocity for height more than 10 thousand kilometer. That made me to think of this question.

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  • $\begingroup$ Escape velocity doesn't work like that. See physics.stackexchange.com/q/29779/123208 Also see en.wikipedia.org/wiki/Specific_orbital_energy & en.wikipedia.org/wiki/Parabolic_trajectory $\endgroup$
    – PM 2Ring
    Commented Nov 26, 2020 at 10:59
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    $\begingroup$ Does this answer your question? How does escape velocity relate to energy and speed? It's not an exact duplicate, but I think the answers there will clear up your misunderstandings. $\endgroup$
    – PM 2Ring
    Commented Nov 26, 2020 at 11:07
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    $\begingroup$ Atul, if we ignore atmospheric resistance, if a body falls back to the planet, it will hit the ground with the same velocity that it was launched with. Conversely, if a body falls to the planet on a linear parabolic trajectory (one with zero specific orbital energy), it will hit the planet with a speed equal to the escape velocity. As those linked Wikipedia articles explain, these results follow from energy conservation. $\endgroup$
    – PM 2Ring
    Commented Nov 26, 2020 at 12:10
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    $\begingroup$ @Atul Those online calculators assume that the acceleration due to gravity $g$ is constant regardless of height $h$. This is a reasonable approximation for values of $h$ that are small compared to the radius of the earth. But Newton's law of universal gravity tells us that it is not true for larger values of $h$. If $g$ really were constant at all values of $h$ then escape velocity would be infinite. $\endgroup$
    – gandalf61
    Commented Nov 26, 2020 at 13:24
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    $\begingroup$ @Atul I have expanded my comment into an answer. $\endgroup$
    – gandalf61
    Commented Nov 26, 2020 at 14:03

2 Answers 2

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If an object travels ballistically (i.e. under the influence of gravity, without any external force or expenditure of energy) and if we neglect air resistance then the kinetic energy the object has when if it returns to its starting point will be the same as the kinetic energy it had to start with. If its mass is unchanged then the magnitude of its final velocity (if it returns to its starting point) must be the same as the magnitude of its initial velocity. This is true if the object is thrown up and comes down, but it also true if it is travelling in an orbit.

As the object travels away from the earth, it loses kinetic energy but gains potential energy. If it is thrown straight up then eventually it may reach the point where all of its kinetic energy has been exchanged for potential energy. This will be the apex of its trajectory, and it will now fall straight down from this apex, exchanging potential energy for kinetic energy on the way. When it reaches its starting point it has the same kinetic energy as it started with.

But if its initial kinetic energy is great enough then the object may never run out of kinetic energy and will continue to travel away from earth for ever - we say it "escapes to infinity", although it obviously cannot reach an infinite distance in finite time. The smallest initial velocity at which this is possible is escape velocity.

The online calculators that you have been using assume that the acceleration due to gravity $g$ is constant regardless of height $h$. This is a reasonable approximation for values of $h$ that are small compared to the radius of the earth. But Newton's law of universal gravity tells us that it is not true for larger values of $h$ - instead $g$ is inversely proportional to the square of the distance from the centre of the earth.

If $g$ really were constant at all values of $h$ then no ballistic trajectories could escape to infinity, and so escape velocity would be infinite

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If you consider only an object and one celestial body, the object does not travel away at a constant speed. The escape velocity is defined as the speed which will take the object an infinite distance away. It is subject to gravity and a loss of speed all the way out. On the way out it loses kinetic energy and gains potential; coming in it loses potential and gains kinetic for any distance traveled.

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