REMARKS Earth\'s motion around the Sun was neglected; that requires using Earth\
ID: 2040911 • Letter: R
Question
REMARKS Earth's motion around the Sun was neglected; that requires using Earth's "sidereal" period (about four minutes shorter). Notice that Earth's mass could be found by substituting the Moon's distance and period into this form of Kepler's third law.
QUESTION If the satellite was placed in an orbit three times farther away, about how long would it take to orbit the Earth once? Answer in days, rounding to one significant figure.
______________ days
EXERCISE
Mars rotates on its axis once every 1.02 days (almost the same as Earth does).
Explanation / Answer
(1) Three times farther away than what?
(2) For "orbit", centripetal accel = gravitational accel
?² r = GM / r²
rearranges to
r³ = GM / ?² = GM / (2?/T)² = GMT² / 4?²
r³ = 6.674e?11 N·m²/kg² * 6.42e23kg * (1.02days * 24hr/day * 3600s/hr)² / 4?²
r³ = 8.43e21 m³
r = 2.04e7 m ? from the center of Mars.
If you need the distance from the surface of Mars, you'll need to look up its radius.
(b) v = ? r = 2?r / T = 2? * 2.04e7m / (1.02days * 24hr/day * 3600s/hr) = 1451 m/s
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