Golfing on the Moon: Can a Ball Achieve Orbit?

  • Thread starter Thread starter Jean-Louis
  • Start date Start date
  • Tags Tags
    Ball Moon Orbit
AI Thread Summary
Hitting a golf ball on the moon to achieve a stable orbit is theoretically possible but requires calculating the necessary exit velocity to overcome lunar gravity. The trajectory of the ball would likely return to the original spot, especially if teed off from a high elevation. Despite the moon's lower gravity, the gravitational force is still significant enough that the ball would not maintain a true orbit. The comparison to launching a ball from Beijing to Tokyo illustrates the challenge of achieving the required distance. Ultimately, while the concept is intriguing, practical execution remains highly improbable.
Jean-Louis
Messages
15
Reaction score
0
Imagine that you are on the surface of the moon, playing golf.

So, is it possible to hit a golf ball on the surface of the moon and have it achieve a stable orbit around the moon?
 
Physics news on Phys.org
You would need to work out the exit velocity required by an object to leave the gravitation force. Then if this is humanly possible.

If so you could
 
The closest you could get to orbit would have a trajectory that would bring it back to where you started. So if you tee'd it up on a mountain, maybe, but otherwise it'd just hit you in the back of the head when it came around again.
 
remember, even though the moon has light gravity, its not as small as you think it is

even with light gravity, it would be like launching the ball from bejing to tokyo
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top