Moon earth system escape velocity

In summary, the question asks for the escape speed of a projectile fired from the far side of the moon, away from the earth. The equation for escape velocity is sqrt[2GM/R], and one must take into account the gravitational potential energy of both the Earth and the projectile. This means that the energies must be added together to determine the projectile's escape speed from the earth-moon system.
  • #1
bcjochim07
374
0

Homework Statement


A projectile is fired straight away from the moon from a base on the far side of the moon, away from the earth. What is the projectile's escape speed from the earth-moon system?


Homework Equations


Escape velocity = sqrt[2GM/R]


The Attempt at a Solution



What I'm wondering is, do I just have to use this formula the moon or do I have to take the gravitational potential energy of the Earth and the projectile into account also?
 
Physics news on Phys.org
  • #2
bcjochim07 said:
What I'm wondering is, do I just have to use this formula the moon or do I have to take the gravitational potential energy of the Earth and the projectile into account also?

Certainly. you need an equation for the energy needed to escape from the Earth and the moon, and you can add those energies.
 
  • #3


I would approach this problem by first considering the factors that affect the escape velocity of an object from the Earth-moon system. These include the masses of the Earth and moon, the distance between them, and the gravitational constant. The formula provided, sqrt[2GM/R], takes all of these factors into account and provides the escape velocity from the system.

In this scenario, since the projectile is being fired away from the moon, we can assume that the distance between the moon and the projectile is negligible compared to the distance between the moon and Earth. Therefore, we can use the mass and radius of the Earth in the formula to calculate the escape velocity.

It is important to note that this formula assumes a point mass for the Earth and moon, which may not be entirely accurate. However, for the purpose of this problem, it provides a reasonable approximation.

In conclusion, the projectile's escape velocity from the Earth-moon system can be calculated using the formula provided, taking into account the mass and radius of the Earth. Other factors, such as the gravitational potential energy, may affect the trajectory of the projectile but are not necessary to calculate the escape velocity.
 

1. What is escape velocity for the Moon-Earth system?

The escape velocity for the Moon-Earth system is the minimum speed an object needs to travel in order to break free from the gravitational pull of both the Moon and the Earth.

2. How is escape velocity calculated?

Escape velocity is calculated using the formula v = √(2GM/R), where G is the universal gravitational constant, M is the mass of the larger body (in this case, the Earth), and R is the distance between the two bodies (the distance from the center of the Earth to the center of the Moon).

3. What is the escape velocity for the Moon-Earth system?

The escape velocity for the Moon-Earth system is approximately 11.2 km/s (kilometers per second).

4. Why is escape velocity important for space travel?

Escape velocity is important for space travel because it determines the amount of energy and speed needed for a spacecraft to break free from the gravitational pull of a celestial body and travel into space.

5. Can escape velocity vary for different objects in the Moon-Earth system?

Yes, escape velocity can vary for different objects in the Moon-Earth system depending on their mass and distance from each other. For example, the escape velocity for the Moon is lower than the escape velocity for the Earth because the Moon has less mass and a weaker gravitational pull.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
2K
Back
Top