## Batting in Space

### Batting in Space

I'm eleven years old and am interested in astronomy and baseball.  I'd
like to know how far a player could hit a baseball on each of the
planets, the sun, and the moon.

A batter has to hit a ball 430 feet to get a home run in Shea Stadium.
Would somebody show me how to calculate how far a ball hit hard enough
to be a home run in Shea Stadium would go on each of the planets, the
moon, and the sun?

--

Justin Florence                   {yale,uunet}!hsi!mlfarm!justin
Owner & General Manager           Maple Lawn Farmers (EFBL)

### Batting in Space

Someone posted a proposed solution for how far Howard Johnson would hit a
ball on the moon, but I think there's a problem with it.

To wit: gravitational attraction diminishes according to an inverse square
law.  The calculation used assumed constant gravitational acceleration, which
is an OK assumption near the surface of the earth, but on something as small
as the moon, you have to start worrying about being able to hit a ball into
orbit.

In addition, the calculation given assumed a flat playfield (planar, that is).
On the moon, even if you don't achieve escape velocity with your dinger, you
might still get a significant fraction of an orbit, making the ball fall not
*to* the surface directly, but *around* the surface.  Sort of like hitting a
ball in Atlanta and having it land in Kuala Lumpur.

Of course, the real answer is "On the moon, *all* stadia are domes". :-)

Are there any astronomers or ballisticians out there who know the appropriate
model for this sort of calculation.  Hell, now *I'm* curious...

--
David M. Tate       | "A fool and his money are soon elected."

| "I was of three minds, like a tree in which
"A Man for all Seasonings" |  there are three blackbirds." -- Wallace Stevens

### Batting in Space

Quote:

>Are there any astronomers or ballisticians out there who know the appropriate
>model for this sort of calculation.  Hell, now *I'm* curious...

The escape velocity from any mass can be written

v = [2gr] **1/2

and g is the acceleration due to gravity on the surface of the mass,

g = GM/(r*r)  .

The escape velocity is thus proportional to the square root of the mass
of the planet and inversely proportional to its radius.  This equation can
be derived simply using newtonian mechanics by taking the zero of
gravitational potential to be at infinity.  Note that this is independent
of the *direction* the ball is hit, either straight "up" or tangentially.

One can also write the acceleration due to gravity as a function of radial
distance above the surface.  To do the calculation, do the 2-D motion
problem from freshman physics assuming a plane surface.  Then look up the
radius of the moon to see if their is significant curvature over the
previously calculated distance-of-flight.  If no, you're done.

If so, one must treat the problem as an elliptical orbit problem in a
non-inertial reference frame, finding the points of intersection between
the ball's path and the lunar surface.

I can't recall the exact data, but I believe the escape velocity of the
moon is too small for it to have an atmoshere.

Have fun!

### Batting in Space

I don't know if it is a good idea to post this, but I just feel like
doing this.

First of all,  you have to know the gravity of all the planets, the
moon, and the sun.  I only know that the gravity of the moon is 1/6 g,
which is a sixth of  the gravity of earth.

First, let me derive an equation.

V(vertical final) = V(vertical initial) - gt

g = gravitational acceleration  t = time

Since the vertical velocity is zero at the top of the trajectory of a
baseball hit by somebody, and I am only talking about half the
trajectory:

V(vertical initial) = gt

I want to know the time the ball takes until it  reaches the ground:

t = 2 (V(vertical initial))/g

I  want to know the horizontal distance the ball takes on our planet.

Horizontal Distance =HD = V(horizontal initial) times time
= 2(V(horizontal initial) V(vertical initial))/g

Now let's assume  that Howard Johnson  can  swing a bat on
any heavenly body so that the initial horizontal velocity and the
initial  vertical velocity of the ball hit by him are always the  same
as those of a 430 foot homerun he hit a few months ago at Shea Stadium

On the moon, the gravity is a sixth of that of our planet.  So:

HD(on the moon) = 2(V(horizontal initial) V(vertical initial)) / (g/6)
=6 (2(V(horizontal initial) V(vertical initial))) / g
= 6 (HD(on earth)) = 6 times 430 feet = 2580 feet

Wow!  Easily a new record.

Since I forgot the gravitational acceleration of each  planet and the
sun, this is all I can do.

The easiest way to know this  is to know that the distance a guy can
hit the ball is inversely  proportional to the gravity of  the place,
assuming  that there is no air friction.

Since I am in hurry,  I have  to go without double-checking.  I  think
this is right  since I am a physics major.

### Batting in Space

Thanks to everybody who replied to my original posting.  From the
formulas and programs that people sent me I can now calculate the
trajectory of a baseball hit on different planets.  I need a little more
help to complete my project.  I'm using a factor of 0.25 for the drag on
a baseball hit on Earth.  With an initial velocity of 180 mph and a
batting angle of 30 degrees, the ball will go 430 feet, which would be a

My question is: what drag factor should I use for other planets?

Thanks again for the help.
--

Justin Florence                   {yale,uunet}!hsi!mlfarm!justin
Owner & General Manager           Maple Lawn Farmers (EFBL)

### Batting in Space

Although I don't have any equations handy to figure out the various
distances, one thing I have noticed about most of the answers- They don't
look right.  Something to do with treating the different gravities as
linear functions.  Doesn't the fact that the object baseball would be
affected by an acceleration force introduce more complexity into the