Asked by Christine
Planet Jupiter revolves around the Sunin 12 years (3.79 X 10^8 sec. What is its mean distance from center of Sun? mass of sun is 1.99 x 10^30 kg.
Answers
Answered by
Damon
F = m A
A = Ac = v^2/R
so m A = Mjupiter v^2/R (toward sun)
F = G Msun MJupiter /R^2 (toward sun)
so
G Msun/R^2 = v^2/R
G Msun = v^2 R
Time around = circumference /v
T = 2 pi R/v
so
v = 2 pi R/T
v^2 = (2pi)^2 R^2/T^2
so
G Msun = (2 pi)^2 R^3/T^2
(which by the way is Kepler's Third Law)
so
R^3 = G Msun T^2/(2 pi)^2
G is 6.67*10^-11
so
R^3 = 6.67*10^-11*1.99*10^30*14.36*10^16 /39.48
R^3 = 4.828*10^35
= .4828 * 10^36
so
R = .784 * 10^12 = 7.84 * 10^11 meters
A = Ac = v^2/R
so m A = Mjupiter v^2/R (toward sun)
F = G Msun MJupiter /R^2 (toward sun)
so
G Msun/R^2 = v^2/R
G Msun = v^2 R
Time around = circumference /v
T = 2 pi R/v
so
v = 2 pi R/T
v^2 = (2pi)^2 R^2/T^2
so
G Msun = (2 pi)^2 R^3/T^2
(which by the way is Kepler's Third Law)
so
R^3 = G Msun T^2/(2 pi)^2
G is 6.67*10^-11
so
R^3 = 6.67*10^-11*1.99*10^30*14.36*10^16 /39.48
R^3 = 4.828*10^35
= .4828 * 10^36
so
R = .784 * 10^12 = 7.84 * 10^11 meters
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