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

In summary, the conversation revolves around the application of Newton's second law F=ma to the moon's gravity of 1.622m/s^2. Various formulas and concepts, such as F=mg and Newton's law of gravitation, are discussed in relation to finding a generalized formula for the moon's gravity. The possibility of using Earth's gravity as a template is also mentioned, but ultimately it is agreed that the mass and radius of the moon are the main factors in determining its gravity.
  • #1
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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  • #2
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?
 
  • #3
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*
 
  • #4
Farhad-.- said:
Objects falling towards the moon*

Simply redefine g. That's all that needs to be done.
 
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  • #5
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 :/
 
  • #6
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.
 
  • #7
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.
 
  • #8
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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  • #9
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 ^•^
 

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

1. What is Newton's second law of motion?

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In simpler terms, the greater the force applied to an object, the greater its acceleration will be, and the more massive the object, the less it will accelerate.

2. How do you calculate force using Newton's second law?

To calculate force using Newton's second law, you would use the formula F=ma, where F is the force in Newtons, m is the mass in kilograms, and a is the acceleration in meters per second squared. This formula can also be rearranged to solve for other variables, such as mass or acceleration, depending on the given information.

3. What is the unit of force in Newton's second law?

The unit of force in Newton's second law is the Newton (N), which is equivalent to kg*m/s². This unit is named after Sir Isaac Newton, the scientist who first described the relationship between force, mass, and acceleration.

4. Can Newton's second law be applied to both stationary and moving objects?

Yes, Newton's second law can be applied to both stationary and moving objects. In the case of stationary objects, the acceleration is equal to 0, so the force acting on the object is also equal to 0. However, for moving objects, the acceleration is not equal to 0, so the force acting on the object can be calculated using the formula F=ma.

5. How does Newton's second law relate to everyday life?

Newton's second law has many practical applications in everyday life. For example, it explains why it is easier to push a shopping cart that is empty rather than one that is full, as the empty cart has less mass and therefore requires less force to accelerate. Additionally, it is the basis for many technologies such as cars, rockets, and elevators, where an understanding of force and acceleration is crucial for their functioning.

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