# A fictitious space probe...

• Mar 15th 2007, 04:41 PM
ceasar_19134
A fictitious space probe...
A ficticious space probe recently discovered a perfectly spherical moon of Jupiter. While taking photographs of the surface, the probe orbited directly above the moon's equator. If the probe's orbit was exactly 2km longer than the circumfrence of the moon, how high above the moon did the probe orbit?
• Mar 15th 2007, 08:53 PM
Soroban
Hello, ceasar_19134!

Quote:

A ficticious space probe recently discovered a perfectly spherical moon of Jupiter.
While taking photos of the surface, the probe orbited directly above the moon's equator.
If the probe's orbit was exactly 2km longer than the circumference of the moon,
how high above the moon did the probe orbit?

It looks like there isn't enough information given in the problem,
. . but the answer is surprisingly simple.

Let R = radius of the moon on kilometers.
Then the circumference of the moon is: 2πR km.

Let h be the height of the probe above the moon's surface.
Then the circumference of the probe's orbit is: 2π(R + h) km.

We are told that two distances differ by 2 km.
. . 2π(R + h) - 2πR .= .2

and we have: .2πR + 2πh - 2πR .= .2 . . 2πh = 2 . . h = 1/π

Therefore, the height of the probe is about 0.3183 kilometers.

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The answer did not depend on the size of Jupiter's moon.

The answer would be same if the probe circled the Sun or a baseball.

• Mar 28th 2007, 03:06 PM
Rimas
and we have: .2πR + 2πh - 2πR .= .2 . . 2πh = 2 . . h = 1/π

Therefore, the height of the probe is about 0.3183 kilometers.

how do you know that ?
• Mar 28th 2007, 05:36 PM
topsquark
Quote:

Originally Posted by Rimas
and we have: .2πR + 2πh - 2πR .= .2 . . 2πh = 2 . . h = 1/π

Therefore, the height of the probe is about 0.3183 kilometers.

how do you know that ?

We are told the probe's orbit is exactly 2 km longer than the circumference of the moon. The circumference of the moon is 2(pi)R. The radius of the orbit of the probe is R + h, where h is the height of the probe over the moon's surface, so the circumference of the circle it traces is 2(pi)(R + h) = 2(pi)R + 2(pi)h. Subtracting the two gives 2 km:
[2(pi)R + 2(pi)h] - 2(pi)R = 2

-Dan