Question

1. two solid spheres of radius r made of the same type of steel are placed in contact, as shown in the figures above. the magnitude of the gravitational force that they exert on each other is f₁. when two other solid spheres of radius 3r made of this steel are placed in contact, what is the magnitude of the gravitational force that they exert on each other? (a) f₁ (b) 3f₁ (c) 9f₁ (d) 81f₁

Explanation:

Step1: Recall the law of universal gravitation

The gravitational force between two objects is given by \( F = G\frac{m_1m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers.

Step2: Analyze the mass of the spheres

The volume of a sphere is \( V=\frac{4}{3}\pi R^3 \). Since the spheres are made of the same steel (density \(
ho \) is constant), mass \( m=
ho V \). For the first sphere with radius \( R \), mass \( m_1=
ho\times\frac{4}{3}\pi R^3 \). For the sphere with radius \( 3R \), mass \( m_2=
ho\times\frac{4}{3}\pi(3R)^3=
ho\times\frac{4}{3}\pi R^3\times27 = 27m_1 \). So each sphere with radius \( 3R \) has a mass 27 times that of the sphere with radius \( R \).

Step3: Analyze the distance between centers

For the two spheres of radius \( R \), the distance between their centers \( r_1 = R + R=2R \). For the two spheres of radius \( 3R \), the distance between their centers \( r_2=3R + 3R = 6R \).

Step4: Calculate the initial force \( F_1 \)

For the first case, \( F_1=G\frac{m\times m}{(2R)^2}=G\frac{m^2}{4R^2} \), where \( m \) is the mass of the sphere with radius \( R \).

Step5: Calculate the force \( F_2 \) for the second case

For the second case, the mass of each sphere is \( 27m \), and the distance between centers is \( 6R \). So \( F_2 = G\frac{(27m)\times(27m)}{(6R)^2}=G\frac{729m^2}{36R^2} \).

Step6: Find the ratio of \( F_2 \) to \( F_1 \)

Divide \( F_2 \) by \( F_1 \): \( \frac{F_2}{F_1}=\frac{G\frac{729m^2}{36R^2}}{G\frac{m^2}{4R^2}}=\frac{729}{36}\times\frac{4}{1}=\frac{729}{9} = 81 \). So \( F_2 = 81F_1 \).

Answer:

D. \( 81F_1 \)