Can a Golf Ball Orbit an Asteroid?

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SUMMARY

The discussion focuses on calculating the size of an asteroid that would allow a golf ball, hit at a speed of 92 m/s, to achieve escape velocity. The calculations involve using the escape velocity formula, \(v_{esc}=\sqrt{\frac{2GM}{R}}\), and the mass of the asteroid derived from its volume and density, \(M=\frac{4}{3} \pi R^3 \rho\), where \(\rho\) is 2.5 g/cm³. The participant successfully determined the radius \(R\) to be 77144.5 m. However, when analyzing the orbit of the golf ball hit at 80 m/s, they encountered an issue resulting in a negative eccentricity, indicating a miscalculation in the orbital parameters.

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  • Understanding of gravitational physics and escape velocity
  • Familiarity with orbital mechanics, including eccentricity and semi-major axis
  • Knowledge of basic calculus for solving equations
  • Ability to manipulate and convert units, particularly density from g/cm³ to kg/m³
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  • Review the derivation of escape velocity and its application in orbital mechanics
  • Learn about the calculation of eccentricity in elliptical orbits
  • Study the relationship between semi-major axis and perihelion distance in orbital dynamics
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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..

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

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 R=a(1-e).

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

v=\sqrt{\frac{m(1+e)}{a(1-e)}}

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 :\
 
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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.
 

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