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The Infinite Drive (Posted on 2006-02-18) Difficulty: 3 of 5
A golf ball cannon is made that shoots the ball at a 45 degree angle to the horizontal, and consistently achieves 300 yards.

That cannon is transported to a spherical asteroid which has the same density as the Earth. A golf ball is fired in a horizontal direction and just reaches the velocity for a stable circular orbit.

What is the radius of the asteroid?
What is the time of one orbit?

Assume the radius of the Earth is 4000 miles.
(Inspired by The Longest Drive)

See The Solution Submitted by Larry    
Rating: 3.5000 (2 votes)

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Solution solution | Comment 2 of 5 |

The universal gravitational constant, G, is 6.673 x 10^-8 cm^3 g^-1 s^-2 and the earth's mass is 5.976 x 10^27 g. Converting G into units of feet and earth masses, we get 1.40827291169018 x 10^16 ft^3 per em s^2. (em indicating earth mass). 

At earth's surface, 4000 mi = 21120000 feet from the center, the acceleration of gravity is then 31.572 ft/s^2.

If v is the muzzle velocity, the ball's maximum height is reached at t = v /  (31.572 * sqrt(2)), but also 2 * t * v / sqrt(2) = 900 feet. The muzzle velocity works out to sqrt(900 * 31.572) ft/s = 168.57 ft/s. This assumes there is no air resistance.

If r is the asteroid's radius relative to the earth, then its mass is r^3 earth masses. However, for distance, we have to use feet, so call the radius in feet, R. Use capital V to represent the orbital velocity sought. Then V^2 = G r^3 / R, but R = 21120000 r, so this is G r^2 / 21120000 (ft/s)^2.

So V^2 = 666795886 r^2 (ft/s)^2.

and V = 25822 r  ft/s

If r were 1, the resulting 25,822 would be the orbital velocity around earth itself.

We want the orbital velocity to equal the muzzle velocity, so

r = 168.57 / 25822 = 0.0065279 earth radii, or 137870 feet or 26.1 miles.

The orbital period is proportional to the 3/2 power of the orbital radius and inversely proportional to the square root of the mass of the body being orbited. So the orbital period is r^(3/2) / sqrt(r^3) = 1 times the period on earth. So the orbital period is the same as the period of a circular orbit around earth at earth's surface.

That orbital period would be 4000 * 2 * pi * 5280 / 25822 seconds, or 1.4275 hours, calculated for earth, but as seen valid for the asteroid.

  Posted by Charlie on 2006-02-18 17:51:00
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