# Unusual escape velocity derivation

• Dilema
In summary, the conversation discusses different approaches to deriving the escape velocity, such as using forces and masses, exploiting time reversal symmetry, or utilizing the conservative nature of the gravitational field. However, it is suggested that these approaches ultimately rely on the principle of energy conservation.
Dilema
Is it possible to derive escape velocity say using momentum and force balance considerations? or using angular momentum consideration?
Namely, any other approach then energy consideration that utilizes gravitation potential energy and kinetic energy?

Dilema said:
Summary: Does anyone knows other derivation then the usual energy conservation to the escape velocity: v=(2MG/r)^0.5?

Is it possible to derive escape velocity say using momentum and force balance considerations? or using angular momentum consideration?
Namely, any other approach then energy consideration that utilizes gravitation potential energy and kinetic energy?
One could, of course, attack the problem from first principles using forces and masses, determining a trajectory and evaluating the required initial velocity to reach a chosen distance using a particular launch angle. One could then take the limit as the target distance is allowed to increase without bound.

Similarly, one could exploit time reversal symmetry and use the same approach to determine the final velocity and impact angle for a drop from a large finite distance with a particular initial angular momentum and then take the limit as the initial distance is allowed to increase without bound.

But it is so much easier to use the fact the the gravitational field is conservative. It then follows immediately that the trajectory is irrelevant and that only the starting and ending points matter.

PeroK and anorlunda
Thanks
jbriggs444
If you found a document that derived it please let me know.

jbriggs444 said:
One could, of course, attack the problem from first principles using forces and masses, determining a trajectory and evaluating the required initial velocity to reach a chosen distance using a particular launch angle. One could then take the limit as the target distance is allowed to increase without bound.

You could. But adding up the forces in tiny little steps looks a lot like integrating the forces in infinitesimal steps, and that's just energy.

I suspect that any derivation falls into this category - it's an obfuscated or at least decorated restatement of energy conservation.

PeroK and jbriggs444
But adding up the forces in tiny little steps looks a lot like integrating the forces in infinitesimal steps, and that's just energy.
Right. It's the process used in calculus-based introductory physics textbooks to derive the expression for the electric potential energy. When one uses conservation of energy to find the escape velocity one is making use of the results derived from integrating the force.

## 1. What is escape velocity and why is it important?

Escape velocity is the minimum speed required for an object to escape the gravitational pull of a planet or other celestial body. It is important because it determines whether an object will remain in orbit or eventually leave the planet's gravitational influence.

## 2. How is the escape velocity calculated?

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

## 3. What factors affect the escape velocity?

The escape velocity is affected by the mass and size of the planet, as well as the distance from the center of the planet. Objects with larger masses or closer distances will have a higher escape velocity.

## 4. Can the escape velocity be exceeded?

Yes, it is possible to exceed the escape velocity, but this would require additional force or propulsion. Objects with a higher velocity than the escape velocity will leave the planet's gravitational influence and enter into space.

## 5. How is the derivation of the escape velocity different for different celestial bodies?

The derivation of the escape velocity is the same for all celestial bodies, but the values for the gravitational constant, mass, and radius will vary depending on the specific planet or object. This means that the escape velocity will be different for each celestial body.

• Mechanics
Replies
5
Views
1K
• Mechanics
Replies
6
Views
2K
• Mechanics
Replies
5
Views
966
• Mechanics
Replies
4
Views
1K
• Mechanics
Replies
13
Views
8K
• Special and General Relativity
Replies
8
Views
552
• Introductory Physics Homework Help
Replies
17
Views
1K
• Mechanics
Replies
4
Views
1K
• Mechanics
Replies
2
Views
932
• Mechanics
Replies
21
Views
4K