Asked by Muskaan
Assuming the earth to be a uniform sphere of radius 6400 km and density 5.5g/c.c. find the value of g on the surface where G=6.67×10
Answers
Answered by
Damon
F = m g = G M m/r^2
actually G = 6.67 * 10^-11
r = 6,400,000 meters = 6.4 * 10^6
rho = 5.5 *10^3 Kg/m^3
M = rho*(4/3) pi r^3 = 23*10^3 r^3
m g = m *6.67*10^-11 *23*10^3 r^3/r^2
g = 6.67*23 *10^-8 (6.4*10^6)
= 983 *10^-2
= 9.83 m/s^2
close enough :)
actually G = 6.67 * 10^-11
r = 6,400,000 meters = 6.4 * 10^6
rho = 5.5 *10^3 Kg/m^3
M = rho*(4/3) pi r^3 = 23*10^3 r^3
m g = m *6.67*10^-11 *23*10^3 r^3/r^2
g = 6.67*23 *10^-8 (6.4*10^6)
= 983 *10^-2
= 9.83 m/s^2
close enough :)
Answered by
bobpursley
if the radius of Earth had been listed as 6378km, it would have been very close.
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