Newton's second law and how to calculate the force on the mo

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Discussion Overview

The discussion revolves around the application of Newton's second law, specifically in the context of calculating the force on objects falling towards the moon. Participants explore how to express gravitational force using the moon's gravity and whether it can be related to Earth's gravity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests using the formula F=mg for objects falling on the moon, proposing a generalized formula F=m(g/6) based on the moon's gravity of 1.622 m/s².
  • Another participant questions the meaning of "applying it to the moon," seeking clarification on whether it refers to objects falling on the moon or the moon-Earth system.
  • Some participants propose simply redefining g for the moon without using Earth's gravity as a template.
  • One participant suggests using Newton's law of gravitation, F=GMm/R², as an alternative approach to calculate gravitational force.
  • Another participant explains how to express the moon's free fall acceleration in terms of Earth's gravity using the relationship between their masses and radii.
  • One participant notes that the formula for gravitational acceleration remains valid regardless of the density variations within a planet, emphasizing the simplicity of the mass and radius relationship.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to calculate force on the moon, with some favoring a redefinition of g and others advocating for the use of Newton's law of gravitation. There is no consensus on a single method or formula.

Contextual Notes

Participants discuss the implications of using Earth's gravity as a reference point and the potential confusion it may cause. There are also considerations regarding the assumptions made in the calculations and the definitions of variables used.

Farhad-.-
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I, earlier were studying Newton's second law F=ma, or rather the free fall of objects, F=mg.
A thought occurred to me, how do I apply this formula to the moon with the Gravity of 1.622m/^2 and came up with F=m(g/6), however, g/6 = 1.63m/s^2

Normally I'd just use F=ma, but assuming people one day would live there, they'd use g = 1.622m/s^2, which I think many would find confusing. So I tried to make a generalized formula. Is there any way to improve upon the formula?
 
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Farhad-.- said:
I, earlier were studying Newton's second law F=ma, or rather the free fall of objects, F=mg.
A thought occurred to me, how do I apply this formula to the moon with the Gravity of 1.622m/^2 and came up with F=m(g/6), however, g/6 = 1.63m/s^2

Normally I'd just use F=ma, but assuming people one day would live there, they'd use g = 1.622m/s^2, which I think many would find confusing. So I tried to make a generalized formula. Is there any way to improve upon the formula?

What do you mean "apply it to the moon"? Do you mean objects falling on the moon, or the moon-Earth system?
 
Student100 said:
What do you mean "apply it to the moon"? Do you mean objects falling on the moon, or the moon-Earth system?
Objects falling towards the moon*
 
Farhad-.- said:
Objects falling towards the moon*

Simply redefine g. That's all that needs to be done.
 
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Student100 said:
Simply redefine g. That's all that needs to be done.
Yeah, but what I was asking is there any way to use Earth's gravity as a template without actually redefining g, which now I think about, I did not specify, I got the formula f=m(g/6). But is there a better way of doing it? My formula skills aren't that great :/
 
Farhad-.- said:
Yeah, but what I was asking is there any way to use Earth's gravity as a template without actually redefining g, which now I think about, I did not specify

Why not use Newton's law of gravitation in that case?

$$F=\frac{Gm_1m_2}{r^2}$$

Although, g in F=mg is just a dummy variable, it doesn't have an intrinsic meaning until we give it one, so I'm not really understanding where you're going.
 
Student100 said:
Why not use Newton's law of gravitation in that case?

$$F=\frac{GMm}{R^2}$$
We are going through that in friday, I have a basic concept of how to use that but not indept unfortunately.
 
Farhad-.- said:
We are going through that in friday, I have a basic concept of how to use that but not indept unfortunately.

It doesn't use Earth as a template either, (not sure why you'd want to), it uses the nature of the system and the gravitational constant.
 
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Free fall acceleration on Earth is just ##g=G\frac{M_{Earth}}{R^2_{Earth}}##, here ##M_{Earth}## is Earth mass, and ##R_{Earth}## - its radius. For Moon you get ##g=G\frac{M_{Moon}}{R^2_{Moon}}##. Now if you want to express free fall acceleration for Moon in terms of Earth g you just need to take relation ##g_{Moon}/g_{Earth}##. So you will get ##g_{Moon}=\frac{M_{Moon}}{M_{Earth}}\frac{R^2_{Earth}}{R^2_{Moon}}g_{Earth}##
 
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  • #10
Domullus said:
Free fall acceleration on Earth is just ##g=G\frac{M_{Earth}}{R^2_{Earth}}##, here ##M_{Earth}## is Earth mass, and ##R_{Earth}## - its radius. For Moon you get just ##g=G\frac{M_{Moon}}{R^2_{Moon}}##. Now if you want to express free fall acceleration for Moon in terms of Earth g you just need to take relation ##g_{Moon}/g_{Earth}##. So you will get ##g_{Moon}=\frac{M_{Moon}}{M_{Earth}}\frac{R^2_{Earth}}{R^2_{Moon}}g_{Earth}##
It's interesting that the formula for the gravitational acceleration at or above the surface is true, even if the density of the planet varies with depth (i.e a massive iron core or shell) It's just the total mass and the radius that count. It makes life a lot simpler :smile:.
 
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  • #11
sophiecentaur said:
It's interesting that the formula for the gravitational acceleration at or above the surface is true, even if the density of the planet varies with depth (i.e a massive iron core or shell) It's just the total mass and the radius that count. It makes life a lot simpler :smile:.
Yeah I know, I had the same thought when I was thinking about it ^•^
 

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