A 12 kg meteor experiences an acceleration of 7.2 m/s^2....

[LionLieOn]

Homework Statement

A 12 kg meteor experiences an acceleration of 7.2 m/s2, when falling towards the earth.

a) How high above the earth’s surface is the meteor?

1.06x10^6

b) What force will a 30 kg meteor experience at the same altitude?

216 N

Homework Equations

I was comparing question b) with a friend and we both got the same answer however, we both had different solutions on solving it. My friend used [Fg= mg] while I used [Fg= Gm1m2/r^2] (m1 is m subscript 1 and m2 is m subscript 2.)

So my question is, does it matter which formula I use?

The Attempt at a Solution

 

[Isaac0427]

No, it doesn't mater, however what you did made the problem more difficult than it needed to be.

Both of those equations are really the same. We know that
$F_g=\frac{Gm_1m_2}{r^2}$
Consider that $m_1$ is the mass of the larger body (earth in this case) and $m_2$ is the mass of the smaller object (the meteor in this case). The force on the meteor is
$m_1\frac{Gm_1}{r^2}=m_1g$
And thus
$g=\frac{Gm_1}{r^2}$
You essentially calculated g yourself, even though it was already given to you.

 

[Cutter Ketch]

No, it doesn't mater, however what you did made the problem more difficult than it needed to be.

Both of those equations are really the same. We know that
$F_g=\frac{Gm_1m_2}{r^2}$
Consider that $m_1$ is the mass of the larger body (earth in this case) and $m_2$ is the mass of the smaller object (the meteor in this case). The force on the meteor is
$m_1\frac{Gm_1}{r^2}=m_1g$
And thus
$g=\frac{Gm_1}{r^2}$
You essentially calculated g yourself, even though it was already given to you.

I'm afraid this reply is flawed. g is not the acceleration due to gravity wherever gravity exists. g is a defined constant = 9.8m/s^2. The two equations are the same only at the surface of the Earth where the radius is the radius of the earth. As stated in the first part the acceleration is NOT g and as calculated in the first part the radius is is NOT the radius of the earth.

What IS true is Newton's second law of motion

F = m a

So that

F = m a = G M m / r^2

The m's cancel and

a = G M / r^2

So if you are clever you note that the acceleration due to gravity is independent of the mass of the object being accelerated everywhere, not just at the surface of the earth. Presuming your friend understood that and noting that a is given in the first part it is valid to write, once again using Newton's second law,

F = m a = 30 kg * 7.2 m/s^2

So it looked to you like something you've been told about gravity: F = m g, but this is only true at the surface of the earth. What he actually used Newton's second law F = m a which is true everywhere.

 

[berkeman]

I used [Fg= Gm1m2/r^2]

This is the correct technique to use away from the surface of the Earth, as pointed out by others.

 

[berkeman]

I'm afraid this reply is flawed

Agreed.

 

[Isaac0427]

Ok. I was actually taught that g was the general term for the gravitational field, and at the surface of the Earth it takes on the value of 9.8 m/s^2. I'm not questioning the experts, I'm just saying that is how I was taught it, which is why I answered that's way. Also, it is on the Wikipedia article https://en.m.wikipedia.org/wiki/Gravitational_field under classical mechanics.

 

[LionLieOn]

Thanks guys! I was a bit confused at first but it makes sense now.

 

[PeroK]

Ok. I was actually taught that g was the general term for the gravitational field, and at the surface of the Earth it takes on the value of 9.8 m/s^2. I'm not questioning the experts, I'm just saying that is how I was taught it, which is why I answered that's way. Also, it is on the Wikipedia article https://en.m.wikipedia.org/wiki/Gravitational_field under classical mechanics.

I'm with you on this. Perhaps you need to be careful using $g$ but I'm not sure that wherever you use it, it must mean the surface gravity of the Earth (which isn't exactly constant in any case). It's not like $G$, which is a universal constant.