Asked by nikki
Ceres, the largest known asteroid, 744 km across, orbits at a mean distance of 4.14 x 10^8 km from the Sun. Find its orbital period.
Answers
Answered by
Anonymous
centripetal acceleration = gravitational acceleration
R omega^2 = G Msun /R^2
R*3 omega^2 = G Msun
(4.14*10^11)^3 omega^2= 6.67*10^-11 *2*10^30 kg
omega^2 = [ 6.67*2 / 4.14^3 ] * 10^[ -11+30 - 33]
omega^2 = 0.167 * 10^-14
omega = 0.408 * 10^-7 = 2 pi f = 2 pi/T
T = 2 pi *10^7 / 0.408 = 15.4 * 10^7 seconds
to get hours divide by 3600 seconds/hr so 15400/3600 *10^4
= 4.28*10^4 hours
to get EARTH days divide by 24
42800/24 = 1783 earth days
for earth years divide by 365
4.9 earth years
R omega^2 = G Msun /R^2
R*3 omega^2 = G Msun
(4.14*10^11)^3 omega^2= 6.67*10^-11 *2*10^30 kg
omega^2 = [ 6.67*2 / 4.14^3 ] * 10^[ -11+30 - 33]
omega^2 = 0.167 * 10^-14
omega = 0.408 * 10^-7 = 2 pi f = 2 pi/T
T = 2 pi *10^7 / 0.408 = 15.4 * 10^7 seconds
to get hours divide by 3600 seconds/hr so 15400/3600 *10^4
= 4.28*10^4 hours
to get EARTH days divide by 24
42800/24 = 1783 earth days
for earth years divide by 365
4.9 earth years
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