How Fast Must You Throw a Baseball to Orbit an Asteroid?

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SUMMARY

The discussion centers on calculating the orbital velocity required to throw a baseball into orbit around a spherical asteroid with a mass of 1.12 x 1017 kg and a radius of 20 km. The correct formula for orbital velocity is derived as Vorbital = sqrt(GM/r), leading to a required speed of 19.3 m/s, rather than the initially calculated 27.3 m/s, which mistakenly used the escape velocity formula. The confusion arises from the distinction between escape velocity and orbital velocity, emphasizing the importance of using the correct equations in gravitational physics.

PREREQUISITES
  • Understanding of gravitational physics and orbital mechanics
  • Familiarity with the equations for potential energy (PE) and kinetic energy (KE)
  • Knowledge of the gravitational constant (G = 6.67 x 10-11 N(m/kg)2)
  • Ability to manipulate algebraic equations and perform square root calculations
NEXT STEPS
  • Study the derivation of the orbital velocity formula Vorbital = sqrt(GM/r)
  • Learn about the differences between escape velocity and orbital velocity
  • Explore the concepts of hyperbolic, parabolic, and elliptical orbits in celestial mechanics
  • Investigate real-world applications of orbital mechanics in space missions
USEFUL FOR

Students and educators in physics, aerospace engineers, and anyone interested in understanding the principles of orbital mechanics and gravitational calculations.

Jesse_1
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Homework Statement


If you stand on the surface of a spherical asteroid of mass 1.12*10^17 kg and radius 20km, how fast must you throw a baseball to put it into orbit at 1.0 m above the surface?

Homework Equations


PE=KE
GMm/r^2 = 1/2mv^2
G = 6,67*10^-11

The Attempt at a Solution


I rearranged the above equation to get the equation for orbital velocity..
Vorbital = sqrt (2GM/r)
Plugged in the numbers
sqrt((2*6.67*10^-11)*(1.12*10^17)/(20000 + 1))

With that, I got 27.3 m/s which doesn't seem completely out there. The problem is the question has the answer listed as 19.3 m/s

I could not figure it out, but after some playing around realized that 19.3 is the answer if the equation is sqrt(GM/r).

Am I using the wrong equation? How does the equation work if there is no 2? I'm so confused. Thanks for any help.
 
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Your method seems to be setting the gravitational potential equal to the kinetic energy. So the net energy is E = 0. In general,
1. E > 0 implies a hyperbolic orbit,
2. E = 0 implies a parabolic orbit,
3. E < 0 implies an elliptical orbit.
A circular orbit falls as the third case. What you could do is simply use the velocity-radius relation for uniform circular motion:
v2/R = a.
 
Jesse_1 said:

Homework Statement


If you stand on the surface of a spherical asteroid of mass 1.12*10^17 kg and radius 20km, how fast must you throw a baseball to put it into orbit at 1.0 m above the surface?

Homework Equations


PE=KE
GMm/r^2 = 1/2mv^2
G = 6,67*10^-11

The Attempt at a Solution


I rearranged the above equation to get the equation for orbital velocity..
Vorbital = sqrt (2GM/r)
Plugged in the numbers
sqrt((2*6.67*10^-11)*(1.12*10^17)/(20000 + 1))

With that, I got 27.3 m/s which doesn't seem completely out there. The problem is the question has the answer listed as 19.3 m/s

I could not figure it out, but after some playing around realized that 19.3 is the answer if the equation is sqrt(GM/r).

Am I using the wrong equation? How does the equation work if there is no 2? I'm so confused. Thanks for any help.

Those two answers approximate the ratio of NASA's Moon Shot speed, and the Orbitting craft Speed. Looks like you have calculated "escape velocity", rather than speed needed to get it to orbit.
 

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