I need help with an equation for: Find the force of attraction between a bowling ball of mass 7.26 kg and the earth if the bowling ball was located at sea level. How does this compare to the weight of the bowling ball as calculated by Fg = mg?

Thank you

Very similar answers.

To find the force of attraction between the bowling ball and the Earth, we can use Newton's law of universal gravitation:

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

Where:
F is the force of attraction,
G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2 / kg^2),
m1 is the mass of one object (the Earth in this case),
m2 is the mass of the other object (the bowling ball in this case),
and r is the distance between the centers of the two objects.

In this case, the mass of the bowling ball is given as 7.26 kg. The mass of the Earth is much larger and is approximately 5.972 × 10^24 kg. The distance between the centers of the Earth and the bowling ball (at sea level) is very close to the Earth's radius, which is about 6,371 km or 6,371,000 meters.

Plugging in the values into the formula, we get:

F = (6.67430 x 10^-11 N m^2 / kg^2 * 7.26 kg * 5.972 × 10^24 kg) / (6,371,000)^2

Simplifying the equation:

F = (4.9058 × 10^14) / 4.058 × 10^13

F ≈ 12.07 N

So, the force of attraction between the bowling ball and the Earth is approximately 12.07 Newtons.

To compare this with the weight of the bowling ball as calculated by Fg = mg, where m is the mass of the bowling ball and g is the acceleration due to gravity, we can plug in the values:

Fg = (7.26 kg * 9.8 m/s^2)

Fg ≈ 71.25 N

Comparing the two values, we can see that the force of attraction between the bowling ball and the Earth is significantly smaller (12.07 N) compared to the weight of the bowling ball (71.25 N) as calculated by Fg = mg. This is because the weight of an object is the force of gravity acting on it, while the force of attraction is the mutual force between two objects. Hence, the weight of an object is the force of attraction specifically between that object and the Earth.