August 29, 2007
TA: Tomoyuki Nakayama Monday, March 29th, 2010
PHY 2048: Physic 1, Discussion Section 3081
Quiz 8 (Homework Set # 11)
Name: UFID:
Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit.
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Assume a planet is a uniform sphere of radius R and mass M that (somehow) has a narrow radial tunnel through its center. Also assume we can position an apple of mass m anywhere along the tunnel or outside the sphere. Let FR be the magnitude of the gravitational force on the apple when it is located at the planet’s surface.
a) How far from the surface is there a point where the magnitude of the gravitational force on the apple is (1/3)FR if we move the apple away from the planet.
Let r be the position of the apple measured from the center of the sphere. Since r>R, the sphere exerts force on the apple as if all of its mass were concentrated at the center. We have
Fr = (1/3)FR ⇒ GMm/r2 = (1/3)GMm/R2 ⇒ r = √(3)R
The distance between the apple and the surface is
h = r – R = (√(3)-1)R = 0.732R
b) What is the magnitude of the gravitational force on the apple if the apple is (1/4)R from the center of the planet. Answer in terms of FR.
Only the part of the sphere which is inside of the apple exerts force on the apple. The mass of the inside part is
Min = (4/3)πρ(R/4)3 = (1/64)×(4/3) πρ(R/4)3 = M/64
Thus the gravitational force is
FR/4 = G(M/64)m/(R/4)2 = 1/4×GMm/R2 = 0.250FR
c) How far from the surface is there a point where the magnitude of the gravitational force on the apple is (1/5)FR if we move the apple into the tunnel?
The mass of the part of the sphere which is inside of the apple is
Min = (4/3)πr3 × ρ = (4/3) πr3 × M/((4/3)πR3)) = M(r/R)3
The condition given in the problem yields
Fr = (1/5)FR ⇒ GM(r/R)3m/r2 = (1/5)GMm/R2 ⇒ r = (1/5)R
Measuring the position from the surface, we obtain
h = R – r = [1 – (1/5)]R = 0.800R
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