Gravitational force between two people and Earth spin rate

[wikidrox]

The first one just confuses me because of the word in the brackets.

1. A 70kg boy stands 1.0 m away from a 60 kg girl. Find the force of attraction (gravitational) between them.

For this would I just sub the values into F=Gm1m2/r^2?

The second question is giving me a lot of problems.

2. How fast would the Earth (r=6400km) have to be spinning befor a 100kg person would exert no force on a scale at the equator? Assume the Earth is a perfect sphere and it's mass is 6.0 * 10^24 kg.

For this I took F=ma and subbed in a=4 (pi)^2 r / T^2. Then I made F=0 and solved for T, but my answer gave me a period of 44 hours. This can't be write since it would have to be spinning faster than 24 hours. Please help.

 

[Doc Al]

The first one just confuses me because of the word in the brackets.

It's just a joke--gravitational attraction versus the non-physics kind.

...For this would I just sub the values into F=Gm1m2/r^2?

Right.

The second question is giving me a lot of problems.

Check out this thread:
https://www.physicsforums.com/showthread.php?t=17606

 

[M98Ranger]

For the first question, you are correct. You can use the formula F=Gm1m2/r^2 to calculate the force of attraction between the two individuals. Just remember to use the correct units for mass (kg) and distance (m).

For the second question, you are on the right track. However, you need to use the formula for centripetal force, which is F=mv^2/r. This is because the person on the equator is moving in a circular motion due to the Earth's rotation. So, the force exerted by the Earth's gravity (F) must be equal to the centripetal force (mv^2/r) in order for there to be no net force on the person.

Substituting the values given, we get:

F=mv^2/r
6.67*10^-11 * 6.0*10^24 * 100 / (6400*1000)^2 = 100*v^2 / 6400*1000
v^2 = 9.8*6400*1000
v = 7900 m/s

This means that the Earth would have to be rotating at a speed of 7900 m/s at the equator for a 100 kg person to not exert any force on a scale. This is equivalent to a rotation period of about 84 minutes, which is much faster than the Earth's actual rotation period of 24 hours. So, your answer of 44 hours is not correct.

I hope this helps! Let me know if you have any other questions.