Meteor Gravitational Field Question

AI Thread Summary
A 12 kg meteor falling towards Earth experiences an acceleration of 7.2 m/s², leading to a calculated altitude of approximately 1.06 x 10^6 m above the Earth's surface. The confusion arises regarding whether to use gravitational force (Fg) or the acceleration (a) in calculations for the radius (r). The user questions if they should divide by Fg or a when applying the gravitational equation. For part b, the force experienced by a 30 kg meteor at the same altitude is correctly calculated as 216 N. Overall, the discussion highlights the complexities of gravitational calculations and the importance of using the correct parameters.
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top