Can a Golf Ball Orbit an Asteroid?

In summary, the problem deals with finding the size of an asteroid that would have a speed of 92 m/s as its escape velocity, and also determining the eccentricity and semi-major axis of an elliptical orbit when a golf ball is hit at a speed of 80 m/s. The equations used in the solution include M=\frac{4}{3} \pi R^3 \rho for mass, v_{esc}=\sqrt{\frac{2GM}{R}} for escape velocity, and v=\sqrt{\frac{m(1+e)}{a(1-e)}} for tangential velocity. However, when solving for the eccentricity, a negative value was obtained and the error remains unclear.
  • #1
Xyius
508
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 :\
 
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  • #2
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.
 

Related to Can a Golf Ball Orbit an Asteroid?

1. What is an orbit around an asteroid?

An orbit around an asteroid refers to the path that an object takes as it revolves around an asteroid in space. This path is determined by the gravitational pull of the asteroid and the speed of the object.

2. How is the orbit around an asteroid calculated?

The orbit around an asteroid is calculated using Newton's law of gravitation, which takes into account the masses of the objects and the distance between them. Scientists also use computer simulations and data gathered from spacecraft to accurately calculate the orbit.

3. What factors affect the orbit around an asteroid?

The orbit around an asteroid can be affected by factors such as the mass and gravitational pull of the asteroid, the speed and trajectory of the object, and any external forces acting on the object, such as other nearby asteroids or planets.

4. How long does it take to orbit around an asteroid?

The time it takes to orbit around an asteroid varies depending on the object's speed and distance from the asteroid. Generally, an orbit around an asteroid can take anywhere from a few hours to several years.

5. Why is studying orbits around asteroids important?

Studying orbits around asteroids can provide valuable information about the composition and structure of these objects. It can also help scientists understand the formation and evolution of our solar system, and even provide insights into potential asteroid mining and impact avoidance strategies.

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