Solving Newton's Gravity Equation: Acceleration & Masses

In summary, Newton's gravity equation states that the force of gravity between two masses is proportional to the mass of the two masses and the square of the distance between them. However, this equation is not symmetrical with respect to gravitational attraction, which means that when an object jumps out of a plane, the force of gravity accelerating the object towards the Earth also accelerates the Earth up towards the object. Additionally, using the correct equations for acceleration, it is found that mass is not a significant factor in determining acceleration. This information is helpful for calculating the acceleration of objects in different situations.
  • #1
napoleonmax
5
0
I was thinking about Newton's gravity equation:
F = G * m(1) *m(2) / r^2. I wanted to find the acceleartion due to gravitational attraction that would occur between two objects of different masses so I used the equation F=ma. Then I ran into difficulty: should I use the sum of the masses to solve this or should I use both masses individually. In other words should my acceleration equation be a= F / m(1) + F / m(2) OR
a = F / [m(1)+m(2)] ? Which one is correct, if any? Clearly the two are not equal (if m(1) is 4 and m(2) is 3 then by the first equation the acceleration will be F/4 + F/3 which is 7F/12 and by the second equation the acceleration is F/7). Another problem is that these equations would allow heavier things to fall faster to the Earth (though only significantly on a large scale). That is, the moon would fall faster toward the Earth than would a car because the change in the force would exede the change in the mass. I would appreciate any help on the matter.
 
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  • #2
F=ma is indeed symmetrical with respect to gravitational attraction. So when you jump out of a plane, the force accelerating you down toward the Earth is also accelerating the Earth up toward you.

And yes, a veeeeery large object would not accelerate at G - G works only for objects which are an insignificant fraction of the mass of the Earth (ie, it works pretty well for anything this side of celestial mechanics).
 
  • #3
Originally posted by napoleonmax
I was thinking about Newton's gravity equation:
F = G * m(1) *m(2) / r^2. I wanted to find the acceleartion due to gravitational attraction that would occur between two objects of different masses so I used the equation F=ma. Then I ran into difficulty: should I use the sum of the masses to solve this or should I use both masses individually. In other words should my acceleration equation be a= F / m(1) + F / m(2) OR
a = F / [m(1)+m(2)] ? Which one is correct, if any?
Neither is correct. The right way is to use a1 = F/m1.
Another problem is that these equations would allow heavier things to fall faster to the Earth (though only significantly on a large scale). That is, the moon would fall faster toward the Earth than would a car because the change in the force would exede the change in the mass.
Using the correct equations, you'll find that acceleration is independent of mass. Thus, if they were at the same distance from the earth, the car and the moon would have the same acceleration. Does this help?
 
  • #4


Originally posted by Doc Al
Using the correct equations, you'll find that acceleration is independent of mass. Thus, if they were at the same distance from the earth, the car and the moon would have the same acceleration. Does this help?
What if you put Jupiter at that distnce from the earth? Will the acceleration still be the same?
 
  • #5


Originally posted by russ_watters
What if you put Jupiter at that distnce from the earth? Will the acceleration still be the same?

Oh no! :smile: Well, if Earth exerts the same gravitational force on Jupiter that Jupiter exerts on Earth...
 
  • #6


Originally posted by russ_watters
What if you put Jupiter at that distnce from the earth? Will the acceleration still be the same?

The acceleration of Jupiter towards the Earth would be the same as a car and the moon, yes. BUT, the acceleration of the Earth towards Jupiter would be substantially larger than than the acceleration of the Earth towards a car!
Therefore, for an observer on Earth, Jupiter would APPEAR to accelerate faster.
 
  • #7


Originally posted by Adrian Baker
The acceleration of Jupiter towards the Earth would be the same as a car and the moon, yes. BUT, the acceleration of the Earth towards Jupiter would be substantially larger than than the acceleration of the Earth towards a car!
Therefore, for an observer on Earth, Jupiter would APPEAR to accelerate faster.

It seems counter-intuitive but it does indeed work out that way.

If you calculate the gravitational force between the Earth and Jupiter,

[tex] F_g = \frac{G M m}{r^2}[/tex]

and then calculate Jupiter's Acceleration toward the earth,

[tex] a = \frac{F}{m}[/tex]

it comes out the same as any other mass, large or small, at the same distance. That is what you are doing when you use the equation,

[tex] a_g = \frac{G M}{r^2}[/tex]

After all,

[tex]m a = \frac{GMm}{r^2}[/tex]
 
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  • #8
Originally posted by napoleonmax
I was thinking about Newton's gravity equation:
F = G * m(1) *m(2) / r^2. I wanted to find the acceleartion due to gravitational attraction that would occur between two objects of different masses so I used the equation F=ma.
I made a web page which gives a similar problem. I used two objects of the same mass though. It is not difficult to extend to different masses. Please see bottom of this web page

http://www.geocities.com/physics_world/gr/grav_field.htm

To find the force on an object simply use the principle of superpostition, i.e. add the forces on the "test" body which result from all the other bodies. You'll find that if the other bodies don't move then the acceleration on the test body is independant on the mass of the test body.
 
  • #9
My point was just that the simplifications you can use for two objects of vastly different mass don't work so well with two objects of similar mass. Whether you drop a bowling ball or a car out of a plane, it works fine to consider the Earth completely stationary (not accelerating toward the plane). But when you drop the moon out of that plane, you need to account for both accelerations and probably pick a better reference frame (at the center of mass of the system for example).
 
  • #10
Originally posted by russ_watters
My point was just that the simplifications you can use for two objects of vastly different mass don't work so well with two objects of similar mass. Whether you drop a bowling ball or a car out of a plane, it works fine to consider the Earth completely stationary (not accelerating toward the plane). But when you drop the moon out of that plane, you need to account for both accelerations and probably pick a better reference frame (at the center of mass of the system for example).

Yes. Excellant point of course. Funny that people rarely aks that question. I was posting a forum in the late 90's called Ask Dr. Neutrino and recall a person asking that very astute question and recall in discust that he was flamed for it.

Independance of acceleration in free-fall demands that the falling object be a test particle, i.e. that it is pointlike and has a mass which may be neglected in comparison to the source of the field. To obtain the answer that you seek you can simply replace the mass of the source with the sum of the masses of the two particles. I.e. suppose the two particles are constrained to move along the x-axis. Then it can be shown that

[tex]A = \frac {G(m_1 + m_2)} {R^2}[/tex]

where A is the relative acceleration of the two particles and R is the distance between them.

For derivation please see
http://www.geocities.com/physics_world/mech/two_accel.htm

Integral - Thanks for showing me that! :-)
 
  • #11
Originally posted by napoleonmax
.. should I use the sum of the masses to solve this ..

Yes. Use the sum of the masses. But note that this applies to a special case of the particles moving along a line.
 
  • #12
Originally posted by russ_watters
Whether you drop a bowling ball or a car out of a plane, it works fine to consider the Earth completely stationary (not accelerating toward the plane). But when you drop the moon out of that plane, you need to account for both accelerations and probably pick a better reference frame (at the center of mass of the system for example).
You are of course correct that when talking about the acceleration with respect to the earth of a massive body, then the acceleration of the Earth must be factored in. I was taking an inertial frame as my reference point (silly me).

For straight line motion, it's trivial (as Arcon points out). My advice to napoleonmax is to keep it simple. Find the accelerations of each using an inertial frame (where F=ma), then add them to get the relative acceleration. This results in napoleonmax's first equation:

Originally posted by napoleonmax
In other words should my acceleration equation be a= F / m(1) + F / m(2)
Yes. That gives the correct relative acceleration.

Edit: included quote by napoleonmax and comment.
 
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  • #13
Originally posted by Doc Al
You are of course correct that when talking about the acceleration with respect to the earth of a massive body, then the acceleration of the Earth must be factored in. I was taking an inertial frame as my reference point (silly me).
Its just that Earth is the reference frame people are used to. When they get a situation where the Earth isn't really able to be considered stationary, it throws them a curve. In some cases, Earth can be considered an inertail frame, in some it can't.
 

1. How do you solve Newton's gravity equation?

The Newton's gravity equation is solved by using the formula F = G * (m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the two objects.

2. What is acceleration in Newton's gravity equation?

The acceleration in Newton's gravity equation is the force of gravity acting on an object due to the presence of another object. It is represented by the variable F in the equation F = ma, where m is the mass of the object and a is the acceleration.

3. How do you calculate the masses in Newton's gravity equation?

The masses in Newton's gravity equation can be calculated by measuring the gravitational force between two objects and the distance between them. The mass of one object can be solved for by rearranging the equation F = G * (m1 * m2) / r^2 to m1 = F * r^2 / (G * m2).

4. What is the significance of the gravitational constant in Newton's gravity equation?

The gravitational constant, denoted by G, is a universal constant that relates the force of gravity to the masses and distance between two objects. It is a fundamental constant in physics and helps in understanding the behavior of objects in the presence of gravity.

5. How does Newton's gravity equation apply to real life scenarios?

Newton's gravity equation is used to explain the motion of objects in the universe, from the movement of planets in our solar system to the trajectory of a ball thrown in the air. It is also used in engineering and space exploration to calculate the gravitational force between objects and plan their orbits.

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