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 The Longest Drive (Posted on 2005-12-26)
Is it possible to hit a golf ball on the surface of the moon and have it achieve a stable orbit around the moon?

 No Solution Yet Submitted by Josh70679 Rating: 3.5000 (2 votes)

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 Final answer | Comment 11 of 16 |

From<o:p></o:p>

http://www.braeunig.us/space/orbmech.htm

Assuming a circular orbit, the orbital velocity, v, is  v=sqrt(GM/r)<o:p></o:p>

G=universal gravitational constant = 6.67e11<o:p></o:p>

M=mass of moon = 7.348e22<o:p></o:p>

R=radius of orbit from center of the moon<o:p></o:p>

All units in kg-m-s<o:p></o:p>

<o:p> </o:p>

Height of orbit (above moon’s surface, r, meters)      Orbital velocity (m/s)<o:p></o:p>

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0<o:p></o:p>

</TD> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 58pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=77 x:num="1679.2780244927592">

1679.27802<o:p></o:p>

</TD></TR> <TR style="HEIGHT: 12.75pt; mso-yfti-irow: 1"> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 48pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=64 x:num>

10000<o:p></o:p>

</TD> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 58pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=77 x:num="1674.467708038336">

1674.46771<o:p></o:p>

</TD></TR> <TR style="HEIGHT: 12.75pt; mso-yfti-irow: 2"> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 48pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=64 x:num>

100000<o:p></o:p>

</TD> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 58pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=77 x:num="1632.9569566274677">

1632.95696<o:p></o:p>

</TD></TR> <TR style="HEIGHT: 12.75pt; mso-yfti-irow: 3"> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 48pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=64 x:num>

500000<o:p></o:p>

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1479.84916<o:p></o:p>

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1000000<o:p></o:p>

</TD> <TD style="BORDER-RIGHT: #e0dfe3; PADDING-RIGHT: 5.4pt; BORDER-TOP: #e0dfe3; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0in; BORDER-LEFT: #e0dfe3; WIDTH: 58pt; PADDING-TOP: 0in; BORDER-BOTTOM: #e0dfe3; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" vAlign=bottom noWrap width=77 x:num="1337.921919279715">

1337.92192<o:p></o:p>

</TD></TR></TBODY></TABLE>

<o:p> </o:p>

From<o:p></o:p>

http://www.homewood.k12.al.us/compsci/projects98/eteam/paper.html on golf ball dynamics<o:p></o:p>

<o:p> </o:p>

Vball = (Vclub*1.67)/(1+mball/mclub)<o:p></o:p>

<o:p> </o:p>

From many sources, Mball=0.045kg (approx. maximum)<o:p></o:p>

Assume a heavy club (gives greater range), mclub=5.0kg<o:p></o:p>

Vclub (from many web sources) for golf pros maxes at about 125mph, assume 200mph ~ 90 m/s<o:p></o:p>

<o:p> </o:p>

Therefore Vball  ~ 150m/s, well short of the required velocity to attain orbit.<o:p></o:p>

<o:p> </o:p>

But, what if the orbit is high enough so that the orbital velocity is only 150 m/s?

<o:p></o:p>

This is not possible, with the following explanation.

As the proposed orbit of the golf ball is higher and higher of the surface of the moon, the initial velocity of the ball, as imparted by the club, must have more and more of a vertical component, such that the vertical component of the velocity of the ball reaches zero just as the ball reaches the proscribed orbital height.  At that point, the ball must still have a horizontal velocity equal to the orbital velocity.  If one does the math, the higher the orbit above the surface, the more energy (and therefore more total velocity = vector sum of vertical + horizontal velocity) the ball must have.  This makes sense, since in rocketry terms,  more fuel is required to get to a higher orbit.<o:p></o:p>

<o:p> </o:p>

Therefore, unless you are superman, with an indestructible club and ball, you cannot “hit” a golf ball into orbit about the moon.<o:p></o:p>

 Posted by Kenny M on 2005-12-28 21:58:40

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