# Baseball thrown into orbit

1. Nov 10, 2014

### Wolfmannm

1. The problem statement, all variables and given/known data
If you were on the top of Mt. Chimborazo in Ecuador (0°S, 80°W, elevation 20,564 ft), with what minimum speed (and in what direction) would you have to throw a baseball horizontally so that it would go into a circular orbit
about the earth? (Neglect air friction. Assume a spherical earth, and assume that no other mountains get in the way.) How long would you have to wait for the ball to come back around?

The following data taken from wikipedia.
Mass of earth Me = 6.673x1024 kg
radius of the earth = 6.268 km
radius of earth and mountain rt= 6.377x107 m
Gravitational constant G=6.673x10-11 Nm/kg

I have found the escape velocity of the ball but I have no idea how to determine the angle it needs to be at to go into orbit. I also have no idea what the minimum height is that would be considered an orbit in order to determine how long the ball will need to travel to return. Any help would be appreciated.

2. Relevant equations
V=√(GMe/rt))

3. The attempt at a solution
V=√(6.673x10-11 Nm/kgx 6.673x1024 kg/6.377x107 m
)
V=7.905x103

2. Nov 10, 2014

### Staff: Mentor

Imagine if you threw the ball out horizontally at the escape velocity what would happen?

3. Nov 11, 2014

### Wolfmannm

I assume it would leave the atmosphere at some point, since earth is a sphere, but wouldn't that be the case at any angle between 0 and 180? Would it continue on after it left the atmosphere or would it go into an orbit?

4. Nov 11, 2014

### Staff: Mentor

The escape velocity or more accurately the escape speed is the minimum speed an object needs in order to not fall back to earth.

So if you throw the baseball out horizontally at escape speed and assume no wind resistance it will not fall to earth.

Instead it will orbit the earth so that some time later it will whizz by you again so prepare to duck...

5. Nov 11, 2014

### Wolfmannm

Okay, I understand that part of it, but what direction would it need to be thrown at? I would think that either east or west would be correct since things don't tend to orbit the poles. Throwing it to the east would be going against earths rotation, and thus would ensure that it returns. While throwing it west would be with its rotation and at a slower speed than the earth rotates, so the ball would not return, the earth would return to the ball.

6. Nov 11, 2014

### Staff: Mentor

You have to consider the velocity addition that happens due to the earths rotation. If you throw it in the same direction as the spin then it will have a speed that higher than the escape velocity right and so will leave its orbit.

7. Nov 11, 2014

### SteamKing

Staff Emeritus
I think you are confusing two different speeds here:

1. Escape velocity is the speed an object needs to attain to break free of a planet's gravity completely. If an object attains escape velocity, it doesn't go into orbit.
2. Orbital velocity is the speed an object needs to attain in order to achieve orbit around a planet. The orbital velocity is significantly lower than the escape velocity.

http://en.wikipedia.org/wiki/Escape_velocity

http://en.wikipedia.org/wiki/Orbital_speed

8. Nov 11, 2014

### Staff: Mentor

My apologies, you are right I did confuse them. I always thought that the minimum escape speed was the orbital speed but now I see that for this case you must divide it by sqrt 2. I just the remember the cannon picture where they keep shooting it faster and faster until it orbits and was taught I think that if it went any faster it would escape.

9. Nov 12, 2014

### Orodruin

Staff Emeritus
If it has a speed in between orbital and escape velocity, it will enter an elliptic orbit. The main issue here is: what is the centripetal acceleration required and what is the gravitational acceleration available?

10. Nov 12, 2014

### collinsmark

In addition to the other good posts, don't forget about the rotation of the Earth. You can use that to your advantage. (There's a good reason the chosen location was on the equator).