Can a Golf Ball Orbit an Asteroid?

  • Thread starter Thread starter Xyius
  • Start date Start date
  • Tags Tags
    Asteroid Orbit
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Xyius
Messages
501
Reaction score
4
Homework Statement
The problem deals with a sports player being able to hit a golf ball at a speed of 92 m/s. The first part says to find the size of an asteroid that would have that speed as an escape velocity.

The second part says that if he hits the ball at 80 m/s, what will the eccentricity be as well as the semi major axis a.
So I did the following..

[tex]v_{esc}=\sqrt{\frac{2GM}{R}}[/tex]
and
[tex]M=\frac{4}{3} \pi R^3 \rho[/tex]

Where ρ=2.5g/cm
Converted rho and plugged everything in and got a value of R to be 77144.5m.

So since the player hits the ball at less than the escape velocity it goes into an elliptic orbit. The position of the ball when it first gets hit is right in the position of closest approach.(Right at the perihelion.)

So that means, from geometry [itex]R=a(1-e)[/itex].

The tangential velocity of the orbit is..
[tex]v=\sqrt{\frac{M}{P}}(1+ecos( \theta))[/tex]
Plugging in for P from geometry and taking theta to be zero...

[tex]v=\sqrt{\frac{m(1+e)}{a(1-e)}}[/tex]

This gives me two equations with two unknowns. When I solve, I get a negative eccentricity! I do not know where I am going wrong :\
 
Physics news on Phys.org
Homework Equations M=\frac{4}{3} \pi R^3 \rhov_{esc}=\sqrt{\frac{2GM}{R}}v=\sqrt{\frac{m(1+e)}{a(1-e)}}The Attempt at a SolutionI have attempted to solve the problem and have gotten a negative eccentricity. I do not know where I am going wrong.