Posted by **Anonymous** on Friday, February 18, 2011 at 1:03pm.

An astronaut on the surface of Mars fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume that the free-fall acceleration on Mars is three eighths that on the Earth.(a) What must be the muzzle speed of the package so that it travels completely around Mars and returns to its original location?

(b) How long does this trip around Mars take?

- physics -
**tchrwill**, Sunday, February 20, 2011 at 10:29am
The circular velocity required to orbit a planet derives from V = sqrt[µ/R] where µ = the gravitational constant of the planet and R = the radius(in feet) of the orbiting body around the planet.

For Mars,

µ = 1.512819x10^15 ft.^3/sec.^2

R = 2111 miles = 11,146,080 feet

Therefore, V = sqrt[1.512819x10^15/11,146,080] = 11,650 ft./sec. = 7,943mph.

The time, or period, to complete an orbit derives from

T = 2(Pi)sqrt[r^3/µ]

...= 2(Pi)sqrt[11,146,080^3/1.512819x10^15] = 6011sec. = 100.18 minutes.

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