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

In summary, the conversation discusses the calculation of the velocity needed to put a baseball into orbit 1.0 m above the surface of a spherical asteroid with a mass of 1.12*10^17 kg and radius of 20 km. The equations used include gravitational potential and kinetic energy, and the final answer for orbital velocity is given as 19.3 m/s. However, the student initially gets an answer of 27.3 m/s, which is the escape velocity. The conversation then discusses the difference between orbital velocity and escape velocity.
  • #1
Jesse_1
1
0

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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  • #2
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.
 
  • #3
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.
 

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

What is the orbital velocity of a baseball?

The orbital velocity of a baseball is the speed at which it travels around a central object, such as the Earth or the Sun, in a circular or elliptical path.

How is the orbital velocity of a baseball calculated?

The orbital velocity of a baseball can be calculated using the formula V = √(GM/R), where V is the orbital velocity, G is the gravitational constant, M is the mass of the central object, and R is the distance between the baseball and the central object.

What factors affect the orbital velocity of a baseball?

The orbital velocity of a baseball is affected by the mass of the central object, the distance between the baseball and the central object, and any external forces acting on the baseball, such as air resistance or gravitational pull from other objects.

How does the orbital velocity of a baseball differ on different planets?

The orbital velocity of a baseball will differ on different planets due to variations in the mass and size of the central object, as well as the distance between the baseball and the central object. For example, the orbital velocity of a baseball on Earth is approximately 29,000 kilometers per hour, while on Mars it is around 21,000 kilometers per hour.

What is the significance of the orbital velocity of a baseball?

The orbital velocity of a baseball is significant as it determines the speed at which the baseball travels in its orbit around a central object. It is also important for understanding the dynamics of celestial bodies and how they interact with each other in space.

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