Asked by NANDU
Find the period time of the moon from the following values,
R=6400km,the distance between the earth and the moon=384000km,g=9.8m/s2
R=6400km,the distance between the earth and the moon=384000km,g=9.8m/s2
Answers
Answered by
drwls
First calculate the speed of the moon in its orbit, V.
V^2/D = g*(R/D)^2
(the value of g at the moon's location)
V^2 = g*R^2/D = 1.045*10^6 m^2/s^2
V = 1022 m/s
V = 2 pi D/P
P (the period) = 2 pi D/V = 2.361*10^6 s
= 27.3 days
Note that D and R had to be in meters, not km, for these calculations
If you need a single equation:
P = 2*pi*D/V = 2*pi*D*sqrt[D/gR^2]
= 2*pi*D^3/2/[R*g^1/2)]
V^2/D = g*(R/D)^2
(the value of g at the moon's location)
V^2 = g*R^2/D = 1.045*10^6 m^2/s^2
V = 1022 m/s
V = 2 pi D/P
P (the period) = 2 pi D/V = 2.361*10^6 s
= 27.3 days
Note that D and R had to be in meters, not km, for these calculations
If you need a single equation:
P = 2*pi*D/V = 2*pi*D*sqrt[D/gR^2]
= 2*pi*D^3/2/[R*g^1/2)]
Answered by
Damon
g of earth at moon = 9.8 (6400/384000)^2
=.00272 m/s^2
so
.00272 = w^2 r = w^2 (3.84*10^8) (note r in meters)
w^2 = 7.1 * 10^-4 * 10^-8
w = 2.66 * 10^-6
w = 2 pi/T
so
T = 2.35 * 10^6 seconds
/3600 s/hr /24 hr/day = 27.3 days
=.00272 m/s^2
so
.00272 = w^2 r = w^2 (3.84*10^8) (note r in meters)
w^2 = 7.1 * 10^-4 * 10^-8
w = 2.66 * 10^-6
w = 2 pi/T
so
T = 2.35 * 10^6 seconds
/3600 s/hr /24 hr/day = 27.3 days
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