Newton’s Law of Gravity specifies the magnitude of the interaction force between

Question

Newton’s Law of Gravity specifies the magnitude of the interaction force between two point masses, m1 and m2, separated by a distance r as F(r) = Gm1m1/r2. The gravitational constant G can be determined by directly measuring the interaction force (gravitational attraction) between two sets of spheres by using the apparatus constructed in the late 18th century by the English scientist Henry Cavendish. This apparatus was a torsion balance consisting of a 6-ft wooden rod suspended from a torsion wire, with a lead sphere having a diameter of 2 in and a weight of 1.61 lb attached to each end. Two 12-in, 348-lb lead balls were located near the smaller balls, about 9 in away, and held in place with a separate suspension system. Today’s accepted value for G is 6.674·10–11 m3kg–1s–2.

a) Determine the force of attraction between the larger and smaller balls at a distance of 6.75 inches between the surfaces that had to be measured by this balance.

b) Compare this force to the weight of the small balls. What is the ratio?

Explanation / Answer

F = ??
G = 6.674 * 10^-11
m1 = 1.61 lb = 1.61/2.2 kg = 0.732 kg
m2 = 348 lb = 348/2.2 = 158.2 kg

for R

we need to convert to meters
r = 4.78 inches + 1 inch + 6 inches = 11.78 inches
r = 11.78 inches *[ 2.54 cm/1 inch] * [1 meter / 100 cm] = 0.299 meters.

Now we need to find F with the equation given
F = [6.674 * 10^-11 * 0.732 kg * 158.2 kg] / 0.299 ^2
F = 8.633*10^-8 N

The force of gravity on the smaller ball is F = m*g
m = 0.732 kg
g = 9.81 m/s^2

F_small_ball = 0.732 * 9.81 = 7.18 N
Ratio = Cavendish Result to small ball = 8.633 * 10^-8 N / 7.18 = 1.20 * 10^-8

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