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.