The centers of a 6.0 lead ball and a 70 lead ball are separated by 8.0 .

What gravitational force does each exert on the other?
What is the ratio of this gravitational force to the weight of the 70 ball?

F=GMm/r^2

To calculate the gravitational force between the two lead balls, you can use Newton's law of universal gravitation:

F = G * ((m1 * m2) / r^2)

Where:
F is the gravitational force
G is the gravitational constant (6.67430 × 10^-11 N m^2/kg^2)
m1 and m2 are the masses of the two lead balls
r is the separation between their centers

First, let's calculate the gravitational force between the two lead balls:

m1 = mass of the 6.0 kg lead ball = 6.0 kg
m2 = mass of the 70 kg lead ball = 70 kg
r = separation between their centers = 8.0 meters

Using the given values:

F = G * ((m1 * m2) / r^2)
F = (6.67430 × 10^-11 N m^2/kg^2) * ((6.0 kg * 70 kg) / (8.0 m)^2)

Now, let's calculate this value:

F = (6.67430 × 10^-11 N m^2/kg^2) * ((6.0 kg * 70 kg) / (8.0 m)^2)
F = (6.67430 × 10^-11 N m^2/kg^2) * ((420 kg^2) / 64 m^2)

Simplifying:

F = (6.67430 × 10^-11 N m^2/kg^2) * (6.5625 × 10^3 kg^2 / m^2)

F = 4.378 × 10^-7 N

So, the gravitational force between the two lead balls is approximately 4.378 × 10^-7 N.

To find the ratio of this gravitational force to the weight of the 70 kg lead ball, we need to calculate the weight of the 70 kg ball:

Weight = mass * acceleration due to gravity

The acceleration due to gravity on Earth is approximately 9.8 m/s^2. Therefore, the weight is:

Weight = 70 kg * 9.8 m/s^2

Weight = 686 N

Now, let's calculate the ratio:

Ratio = Gravitational Force / Weight

Ratio = (4.378 × 10^-7 N) / (686 N)

Ratio ≈ 6.39 × 10^-10

Therefore, the ratio of the gravitational force between the two lead balls to the weight of the 70 kg ball is approximately 6.39 × 10^-10.