Meteor Gravitational Field Question

Click For Summary
SUMMARY

The discussion revolves around calculating the altitude of a meteor and the force experienced by another meteor at that altitude. A 12 kg meteor experiences an acceleration of 7.2 m/s², leading to an altitude of approximately 1.06 x 10⁶ m above the Earth's surface. The force experienced by a 30 kg meteor at the same altitude is calculated to be 216 N. The confusion arises from whether to use gravitational force (Fg) or the acceleration (a) in the calculations, highlighting the importance of understanding the relationship between mass, acceleration, and gravitational force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with gravitational force calculations using Fg = Gm1m2/r²
  • Knowledge of basic physics concepts such as mass, acceleration, and force
  • Ability to manipulate equations involving square roots and constants like G (6.67 x 10^-11)
NEXT STEPS
  • Study the derivation and application of Newton's law of universal gravitation
  • Learn how to apply the formula Fg = ma in different contexts
  • Explore gravitational acceleration variations at different altitudes
  • Investigate the implications of mass and acceleration in real-world scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and gravitational forces, as well as educators looking for examples of gravitational calculations in action.

Ess_Elle
Messages
2
Reaction score
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
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.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
13K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K