Week 2 Lecture 6
Solutions at the bottom.

1)  In-video quiz at timepoint 08:51.

Ronen mentioned that Henry Cavendish measured Newton's constant to be G = 6.67 e-11 Nm²/Kg². This occurred in 1798 in Britain, when Cavendish measured the force of attraction between two lead balls, one small sphere with a mass of 0.73 kilograms and a larger sphere with a mass of 158 kilograms. These were separated by a distance of 0.23 meters.

Use Newton's formula for gravitational force, F = Gm₁m₂/R₂, to determine how much force was exerted by each ball on one another. Give your answer in Newtons using two significant figures.

[Open-end]

Solutions:

1) The correct answer is [1.4e-7,1.6e-7]
This is a straightforward entry of values into our formula, F = Gm₁m₂/R₂. 
G = 6.67 e-11 Nm²/Kg², 
m₁ = 0.73 kg, m₂ = 158 kg, 
and R = 0.23 m.

Plugging in these values, we get F = (6.67e-11)*(0.73)*(158)}/0.23², without any need to convert units, because they all match.

The result is F = 1.45428... e-7 Newtons, which would round to 1.5 e-7 Newtons for our final answer.

For comparison, consider that this force is roughly equivalent to the weight of a grain of sand with a mass only a fraction of a milligram. Remarkable sensitivity!