Quick easy graviational problem

  • Thread starter Demin
  • Start date
In summary, to calculate how fast the Earth would need to spin in order for a 75kg person at the equator to be weightless, we can use the equation GMm/r^2 - N = m*v^2/r and set the normal force (N) equal to zero. This means that the Earth would need to spin fast enough for the person's feet to just leave the ground.
  • #1
Demin
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0

Homework Statement



How fast would the earth(Massearth =5.98X10^24 KG, Radius Earth = 6.4X10^6 m) have to spin on it axis so a 75kg person at the equator would be weightless.

Homework Equations



T^2= kr^3
Fg = G m1 m2 / r^2


The Attempt at a Solution



i know that first i need 2 find the W or the person then put the weight equal to some equation that i don't know =/ .. some 1 please help me ;P
 
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  • #2
Perform a force balance for a person standing on the equator and set that equal to the centripedal acceleration. Next, what does it mean to be weightless?
 
  • #3
dwintz02 said:
Perform a force balance for a person standing on the equator and set that equal to the centripedal acceleration. Next, what does it mean to be weightless?

weightless = u weight 0 N ;/ i still haven't figured it out .. do u mean

(G)(75kg)(5.98X10^24)/6.4X10^6 = V^2/r


? please respond
 
  • #4
For a person standing on the Earth at the equator, how about:

GMm/r^2 - N = m*v^2/r

Sorry for the confusing post earlier, I meant mass times the centripedal acceleration. Remember the normal force (N) is still relevant here because the Earth is physically pushing the person away from the center. Ok now, which of the terms in my equation should go to zero as the Earth spins faster and faster (person becomes more and more weightless)? Or, if it helps, imagine the Earth is spinning just fast enough and the person's feet JUST leave the ground, then what term will be zero?
 

1. How do I solve a quick and easy gravitational problem?

To solve a quick and easy gravitational problem, you will need to know the mass of the objects involved and the distance between them. You can then use the formula F = G(m1m2)/r^2 to calculate the force of gravity between the objects. G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2).

2. What is the difference between weight and mass in gravitational problems?

Weight is a measure of the force of gravity on an object, while mass is a measure of the amount of matter in an object. In gravitational problems, weight is calculated by multiplying mass by the acceleration due to gravity (9.8 m/s^2 on Earth).

3. How does distance affect gravitational force?

According to the inverse square law, as the distance between two objects increases, the force of gravity between them decreases. This means that the closer two objects are, the stronger the gravitational force between them will be.

4. Can I use the same formula for any two objects in a gravitational problem?

Yes, the formula F = G(m1m2)/r^2 can be used for any two objects in a gravitational problem. However, keep in mind that the value of G may vary depending on the units used (e.g. kg or g).

5. Is there a limit to the distance at which gravity can act?

No, gravity is a fundamental force that acts between any two objects with mass, regardless of the distance between them. However, the force becomes weaker as distance increases, making it practically negligible at very large distances.

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