Orbital velocity of a baseball

[Jesse_1]

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.

 

[Gear300]

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.

 

[PeterO]

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.