.11 Sprint 8:08 PM Exit What is the gravitational force of Moon on Earth? FM on
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Question
.11 Sprint 8:08 PM Exit What is the gravitational force of Moon on Earth? FM on E 1.99 x10 20 Show your answer with 3 significant figures. Question 8 2 pts The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 5.0 earth years. Assume a 365.25-days year and Msun-1.99x1030 kg. (a) What is the asteroid's orbital radius? Show your answer with 2 significant figures. (b) What is the asteroid's orbital speed? Show your answer with 2 significant figures. x10 x10 m/s D Question 9 2 pts
Explanation / Answer
8] a] By using Keplers third law,
T^2 = 4*pi^2*r^3/GM
r^3 = T^2*GM/[4*pi^2]
Where, G is the universal gravitational constant,
M is the mass of the sun,
T is the asteroid's period in seconds, and
r is the radius of the orbit.
Covert 5.00 years to seconds
5.00years = 5.00years*(365.25days/year)*(24hours/day)*(60min/hours)*(60 sec/min)
= 1.58 * 10^8 s
Now, the radius of orbit,
r = cuberoot[(6.67*10^-11)*(1.99 * 10^30)*(1.58 *10^8)^2 / (4*pi^2)]
= 4.38*10^11 m
b] The orbital speed can be found from the circular velocity formula
v = sqrt[GM/r]
= sqrt[(6.67*10^-11)*(1.99 *10^30) / (4.38*10^11)]
= 1.74*10^4 m/s
7] F = GMm/r^2
Where, G = Gravitational constant
M = mass of the Earth
m = mass of the Sun
r = distance between the Earth and the Sun
F = 6.674e-11*5.97e24*1.99e30/[1.50e11]^2
= 3.52 * 10^22 N answer
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