Question

In one of his novels author Jules Verne imagined that astronauts inside a spaceship walked on the floor of the cabin when the force exerted on the ship by Earth was greater than the force exerted by the Moon. When the force exerted by the Moon was greater, he thought the astronauts walked on the ceiling of the cabin. (a) At what distance from the center of Earth would the forces exerted on the spaceship by Earth and the Moon be equal? (b) Explain why Verne's description of gravitational effects is incorrect.

Solution

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Concept:\textbf{Concept:}

By applying the Newton’s universal law of gravitation we find the attractive forces from Earth and the Moon. Let the force applied to Earth equal to the force due to the Moon when the spaceship is at a separation rr from the center of Earth. Let R=3.84×108mR=3.84\times 10^8m, which is the distance between the centers of Earth and the Moon. Lastly solve the expression for the distance rr.

Solution:\textbf{Solution:}

Use the assumption that FE=FMF_E=F_M and use law of gravity to solve for rr we get

GmSMEr2=GmSMM(Rr)2\cancel{G}\frac{\cancel{m}_SM_E}{r^2}=\cancel{G}\frac{\cancel{m}_SM_M}{(R-r)^2}

mS(Rr)2=mMr2Rr=2mM/2ErR=2mM/mEr+rm_S(R-r)^2=m_M r^2\Rightarrow R-r =\sqrt{2m_M/2_Er} \Rightarrow R = \sqrt{2m_M/m_Er}+r

r=R1+mM/mE=3.84×108m1+(7.35×1022kg)/(5.97×1024kg)=3.46×108mr=\frac{R}{1+\sqrt{m_M/m_E}}=\frac{3.84 \times 10^8m}{1+\sqrt{(7.35 \times 10^{22}kg)/(5.97 \times 10^{24}kg)}}=\boxed{ \bf \color{#4257b2}3.46 \times 10^8m}

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