# Throwing a baseball while standing on an Asteroid

• takelgith
In summary, the conversation discusses an asteroid with a radius of r and density of 2100 kg/m^3. The question is posed about throwing a baseball at a speed of 28 m/s and determining the largest radius asteroid where the baseball can travel in a circular orbit. Using the equation GMm/r^2 = mv^2/r and substituting the density to find M, the result is r = √[(28^2)/(2100π)(6.67^-11)]. However, upon further discussion, it is discovered that the correct formula should be M = 2100 *(4/3 pi r^3).
takelgith
Summary:: Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around the Sun. Consider an asteroid that is spherically shaped with radius
r and density 2100 kg/m^3.

a.
You find yourself on the surface of this asteroid and throw a baseball at a speed of 28
m/s. If the baseball is to travel around the asteroid in a circular orbit, what is the largest radius asteroid on which you are capable of accomplishing this feat??

- For this I equated GMm/r^2 = mv^2/r and solved for r. I used the density to find M.
I got r = √[(28^2)/(2100π)(6.67^-11)]
it says its wrong!

Last edited by a moderator:
takelgith said:
- For this I equated GMm/r^2 = mv^2/r and solved for r. I used the density to find M.
I got r = √[(28^2)/(2100π)(6.67^-11)]
it says its wrong!
Can you elaborate on the part where you used the density to find M?

jbriggs444 said:
Can you elaborate on the part where you used the density to find M? You did it wrong.

So I had M = 2100π R^3.

Then I subbed that into the equation, for which the R got canceled and became R^2 and thus my result.

takelgith said:
So I had M = 2100π R^3
Where does that formula come from?

takelgith said:
So I had M = 2100π R^3.

Then I subbed that into the equation, for which the R got canceled and became R^2 and thus my result.

Oh shoot! Its supposed to be 4/3piR^3. So M= 2100 *(4/3 pi r^3) right? that should be right??

takelgith said:
Oh shoot! Its supposed to be 4/3piR^3. So M= 2100 *(4/3 pi r^3) right? that should be right??
I think so. I did not spot any other errors.

takelgith

## 1. How far can a baseball be thrown on an asteroid?

The distance a baseball can be thrown on an asteroid depends on several factors, such as the size and gravity of the asteroid, as well as the strength and technique of the person throwing it. However, it is safe to say that a baseball can be thrown much farther on an asteroid than on Earth due to the lower gravity and lack of air resistance.

## 2. Will the trajectory of a thrown baseball be affected by the lack of air on an asteroid?

Yes, the trajectory of a thrown baseball will be affected by the lack of air on an asteroid. On Earth, air resistance plays a significant role in the trajectory of a thrown object. However, on an asteroid with little to no atmosphere, there is no air resistance to slow down or alter the path of the baseball.

## 3. How does the gravity on an asteroid affect the speed of a thrown baseball?

The gravity on an asteroid will affect the speed of a thrown baseball, just like on Earth. However, since the gravity on an asteroid is much weaker than on Earth, the baseball will travel at a much faster speed and cover a greater distance before falling to the ground.

## 4. Can a baseball be thrown in a straight line on an asteroid?

Yes, a baseball can be thrown in a straight line on an asteroid, just like on Earth. However, due to the lower gravity and lack of air resistance, the baseball may follow a slightly curved path before reaching its target.

## 5. Is it possible for a baseball to escape the gravitational pull of an asteroid when thrown?

No, it is not possible for a baseball to escape the gravitational pull of an asteroid when thrown. The gravitational force of an asteroid is still strong enough to keep the baseball within its orbit, even if it is thrown at a high speed. Only objects with a much greater speed, such as a spacecraft, can escape the gravitational pull of an asteroid.

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