Meteor Gravitational Field Question

In summary, when a 12 kg meteor is falling towards the earth, it experiences an acceleration of 7.2 m/s^2. To find the altitude above the earth's surface, the equation r = √(Gm1m2/Fg) can be used, where G is the gravitational constant, m1 and m2 are the masses of the meteor and the earth, and Fg is the force of gravity. When using this equation, the force of gravity can be found using Fg = ma, where m is the mass of the meteor and a is the acceleration experienced. Using this method, the altitude can be determined to be 1.06 x 106 m. For a 30 kg meteor at
  • #1
Ess_Elle
2
0

Homework Statement


A 12 kg meteor experiences an acceleration of 7.2 m/s^2 when falling towards the earth.
a) How high above the Earth's surface is the meteor?
b) What force will a 30 kg meteor experience at the same altitude?

Homework Equations



F
g= Gm1m2/r2 m1m2

r2 = Gm1m2/Fg

r = √Gm1m2/Fg

The Attempt at a Solution



Please help, I have been stuck on this question for hours trying to figure out if what I am doing is right. I am really confused right now, but here is my attempt at the solution:

Fg= Gm1m2/r2 m1m2

r2 = Gm1m2/Fg

r = √Gm1m2/Fg

r = √(6.67x10^11)(5.98x10^24)/7.2 m/s

r = 7442987.005 = 7.44 x 106 m

Paraphrase: The altitude above the earth’s surface is 7.44 x 106 m - 6.38 x 106 m = 1060000 = 1.06 x 106 m.

Now, I am confused because in an example in my lesson, they divide the entire sq. root equation by the force of gravity found, rather than by acceleration (which in this case is 7.2 m/s). I am unsure if I am supposed to divide by Fg rather than a? See below what I mean:

Fg = mg
Fg = (12 kg)(9.8 N/kg)
Fg = 117.6 = 118 N

which would make the equation, r = sq.rt| (6.67 x10^-11)(5.98x10^24)/118 N?

This again confuses me even more, because in this question, I think that Fg = ma = Gm1m2/r^2. So maybe if I am to be dividing by force of gravity, that means, rather than multiplying mg, I would have to multiply ma? See below:

Fg = ma
Fg = (12 kg)(7.2 m/s)

Fg = 86.4 N

If the above equation is correct, then would I have to divide r = sq.rt| (6.67 x10^-11)(5.98x10^24)/86.4 N?


Please help, I am so confused! Am I somewhat on the right track? :/
 
Physics news on Phys.org
  • #2
Oh, I forgot to post my solution to part b). I think I have done part b) correctly. See below:

F = ma

F = (30 kg)(7.2 m/s)

F = 216 N

Paraphrase: Therefore, a 30 kg meteor at the same altitude will experience a force of 216 N.
 

What is a meteor gravitational field?

A meteor gravitational field is a region in space where the force of gravity is affected by the presence of a meteor. This force of gravity can pull objects towards the meteor or alter their trajectory.

How is the strength of a meteor gravitational field determined?

The strength of a meteor gravitational field is determined by the mass and velocity of the meteor. The larger and faster the meteor, the stronger its gravitational field will be.

What happens to objects that enter a meteor gravitational field?

Objects that enter a meteor gravitational field will experience a change in their trajectory. They may be pulled towards the meteor or have their path altered due to the gravitational force.

Can a meteor gravitational field affect Earth?

Yes, a meteor gravitational field can affect Earth if the meteor is large enough and passes close enough to our planet. This can result in changes to Earth's orbit or even impact events.

How does a meteor gravitational field impact space exploration?

A meteor gravitational field can impact space exploration by altering the trajectory of spacecraft and potentially causing collisions. It can also be used to slingshot spacecraft for faster travel through the solar system.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
12K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
955
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
736
  • Introductory Physics Homework Help
Replies
3
Views
900
Back
Top