Can the escape velocity be achieved at any angle?

In summary, the minimum escape velocity is given by the equation vesc = sqrt(2GM/r) and it does not depend on the angle of the velocity vector. It must be directed radially outwards for the object to escape the planet's gravitational pull.
  • #1
nhmllr
185
1
So I looked at a neat derivation for what the minimun escape velocity is, and It was pretty clever. Because this is conserved:
KE + PE = 1/2 *mv^2 + -GMm/r
You can find what the velocity would have to be to get to infinity with zero velocity
1/2 *mvesc^2 + -GMm/r = 1/2 *m(0)^2 + -GMm/(infinity) = 0
1/2 *[STRIKE]m[/STRIKE]vesc^2 = GM[STRIKE]m[/STRIKE]/r
vesc = sqrt(2GM/r)
However, there is no "angle" term included in this. Does this mean that it can travel at this velocity at any angle and it will escape? What if the velocity vector is pointed right at the thing it's orbiting?

Thanks
 
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  • #2
I believe that velocity has to be directed radially outwards. The escape velocity is defined as the velocity required for an object to "get to infinity" with zero remnant velocity. If the velocity is not radially, then that cannot happen from the conservation of angular momentum.
 
  • #3
Yuqing said:
I believe that velocity has to be directed radially outwards. The escape velocity is defined as the velocity required for an object to "get to infinity" with zero remnant velocity. If the velocity is not radially, then that cannot happen from the conservation of angular momentum.

Nope - it works just fine regardless of angle. Keep in mind that from a conservation of angular momentum point of view, v can still go to zero (as r -> inf).
 
  • #4
cjl said:
Nope - it works just fine regardless of angle. Keep in mind that from a conservation of angular momentum point of view, v can still go to zero (as r -> inf).

But mathematically, if you pointed the velocity vector straight at the object with mass M, the force would be infinite when r = 0, and the object would be stationary. Right? Or does the velocity also grow to be near infinite, kind of making them "cancel out?"
 
  • #5
cjl said:
Nope - it works just fine regardless of angle. Keep in mind that from a conservation of angular momentum point of view, v can still go to zero (as r -> inf).

Ah yes, forgot about the fact that r -> inf.

@nhmllr It doesn't really make physical sense to point the velocity towards the object. Escape velocity is defined in terms of the object "escaping" the planet.
 

1. What is escape velocity?

Escape velocity is the minimum speed an object needs to achieve in order to break free from the gravitational pull of a larger body, such as a planet or a star.

2. How is escape velocity calculated?

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 between the object and the center of the larger body.

3. Does escape velocity depend on the angle of launch?

Yes, the angle of launch does affect the escape velocity. The most efficient angle for achieving escape velocity is directly away from the center of the larger body, or at a 90-degree angle. Launching at a shallower angle will require a higher speed, while launching at a steeper angle will require a lower speed.

4. What happens if an object does not reach escape velocity?

If an object does not reach escape velocity, it will remain in orbit around the larger body. It will continue to revolve around the larger body at a constant speed, without ever breaking free from its gravitational pull.

5. Can escape velocity be exceeded?

Yes, it is possible for an object to exceed escape velocity. This typically occurs when an object is propelled by a force, such as a rocket engine, which provides an additional boost of speed. When this happens, the object will break free from the gravitational pull and continue to travel through space at a higher speed.

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