What is the time taken to reach a given height with escape velocity?

In summary: But if you start at the surface of the earth, then the object has lost a lot of its energy due to the resistance of the air, and so the integral becomes more difficult. In summary, an object launched at escape velocity will reach a height h in about 8.3 seconds.
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Gazarai
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If an object is launched at escape velocity from the surface of the Earth, how long will it take to reach a given height h? Ignoring air resistance and other gravity, etc.
 
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That sounds like a pretty basic problem but it isn't, quite. For motion near the surface of the earth, we could take the acceleration due to gravity to be constant, -g. But if you are talking about "escape velocity" you must be considering the object moving to great distance compared to the radius of the earth. Then [itex]F= -GmM/r^2[/itex] where G is the "universal gravitational constant", m is the mass of the object, M is the mass of the earth, and r is the distance from the center of the earth, so that the acceleration is [itex]-GM/r^2[/itex].

That gives [itex]a= dv/dt= -GM/r^2[/itex]. Since the variables on the left are "v" and "t" while we have "r" on the right (and v= dr/dt), we can use a technique called "quadrature". Since a= dv/dt= (dv/dr)(dr/dt) (by the "chain rule") [itex]a= (dv/dr)(dr/dt)= (dv/dr)v= -GM/r^2[/itex]. We can write that as [itex]v dv= (-GM/r^2)dr[/itex].


Integrating [itex](1/2)v^2= GM/r+ C[/itex] (That, by the way, is "conservation of energy"- the left is kinetic energy, the right is the negative of the potential energy, and there difference is constant). Then we have [itex]v^2= 2GM/r+ C[/itex] so that [itex]v= dr/dt= \sqrt{2GM/r+ C}[/itex].

Now, we need to integrate [itex]\frac{dr}{\sqrt{2GM/r+ C}}= dt[/itex].

The left side is an "elliptic integral" (from the fact that it is related to calculating the orbits of planets and satellites which are ellipses), well known NOT to be integrable in terms of elementary functions but well tabulated.
 
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  • #3
If you start at scape velocity, than the constant C vanishes (Which is just the total mechanical energy) vanishes and the integral can be easily performed.
 

1. What is escape velocity?

Escape velocity is the minimum speed needed for an object to leave the gravitational influence of a larger body, such as a planet or moon.

2. How is escape velocity calculated?

The escape velocity can be calculated using the formula v = √(2GM/r), where v is the escape velocity, G is the gravitational constant, M is the mass of the larger body, and r is the distance from the center of the larger body to the object.

3. What factors affect escape velocity?

The escape velocity is affected by the mass and size of the larger body, as well as the distance from the center of the body to the object. A larger and more massive body will have a higher escape velocity, while a smaller and less massive body will have a lower escape velocity.

4. Can escape velocity be exceeded?

Yes, it is possible for an object to exceed the escape velocity of a larger body. This will result in the object leaving the gravitational influence of the larger body and entering into space.

5. How does escape velocity relate to time?

The time it takes for an object to reach escape velocity depends on its initial speed and the acceleration due to gravity of the larger body. The greater the initial speed, the shorter the time it takes to reach escape velocity. Once escape velocity is reached, the object will continue to move away from the larger body at a constant speed unless acted upon by another force.

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