Question #193522

 Q4- The Moon orbits the Earth along a path of radius 3.84 x 108 m. The mass of the Moon is 7.35 x 1022 kg, and the mass of the Earth is 6.00 x 1024 kg. (AC 1.2) a) The gravitational force of attraction between the Earth and the Moon keeps the Moon in orbit. Calculate the size of this force. b) A rocket of mass 42000kg is fired from Earth to the Moon. What is the net gravitational force on the rocket when it is 3 x 108 m from the centre of the Earth?


1
Expert's answer
2021-05-16T17:58:06-0400

Answer

(A)

Gravitational force can be given as

F=Gm1m2r2F=\frac{Gm_1m_2}{r^2}



F=(6.67×1011)(7.35×1022)(6.0×1024)(3.84×108)2F=\frac{(6.67\times10^{-11})(7.35\times10^{22})(6.0\times10^{24})}{(3.84\times10^8)^2}


F=2×1020NF=2\times10^{20}N



(B)

Force on rocket due to earth

F1=(6.67×1011)(42000)(6.0×1024)(3×108)2=186.79NF_1=\frac{(6.67\times10^{-11})(42000)(6.0\times10^{24})}{(3\times10^8)^2}=186.79N


Force on rocket due to moon


F2=(6.67×1011)(42000)(7.35×1022)(0.84×108)2=24.51NF_2=\frac{(6.67\times10^{-11})(42000)(7.35\times10^{22})}{(0.84\times10^8)^2}=24.51N


Net force

Fnet=186.7924.51=162.28NF_{net}=186.79-24.51=162.28N


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS