- #1

ac7597

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- Homework Statement
- Henry Cavendish succeeded in measuring the value of the constant "G" way back in the late 1700s. His method was to put two known masses at a known distance and measure the attractive force between them; then he could use Newton's Law of Universal Gravitation to find "G".

The basic idea was to put a big sphere of a dense material next to a small sphere of dense material. Cavendish chose lead. He made one big ball with radius 5.6 inches, and one small ball with radius 1.9 inches. He then placed them next to each other, with a gap of just 0.5 inches between their closest surfaces.

How large was the attractive force of gravity between these two balls?

- Relevant Equations
- F=G(mass1)(mass2)/(radius)^2

G=6.67 * 10 ^(-11)

**Homework Statement:**Henry Cavendish succeeded in measuring the value of the constant "G" way back in the late 1700s. His method was to put two known masses at a known distance and measure the attractive force between them; then he could use Newton's Law of Universal Gravitation to find "G".

The basic idea was to put a big sphere of a dense material next to a small sphere of dense material. Cavendish chose lead. He made one big ball with radius 5.6 inches, and one small ball with radius 1.9 inches. He then placed them next to each other, with a gap of just 0.5 inches between their closest surfaces.

How large was the attractive force of gravity between these two balls?

**Homework Equations:**F=G(mass1)(mass2)/(radius)^2

G=6.67 * 10 ^(-11)

total radius=((5.6in)+(1.9in)+(0.5in))(0.0254m) = 0.203m

(mass1)(9.8m/s^2)=(6.67 * 10 ^(-11) ) (mass1)(mass2)/( (5.6in)(0.0254 m) )^2

mass2=2.97 * 10^(9) kg

(mass2)(9.8m/s^2)=(6.67 * 10 ^(-11) ) (mass1)(mass2)/( (1.9in)(0.0254 m) )^2

mass1=342.2 * 10^(6) kg

F=(6.67 * 10 ^(-11) )(342.2 * 10^(6)) (2.97 * 10^(9)) / (0.203m)^2

F=1.645 * 10^(9) N

Apparently this incorrect. I don't know why.