Where does the equation for gravitational force come from?

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In summary, the equation {Gmm_e}/{r^2}={mgR^2/r^2} can be derived from Newton's second law and the definition of gravitational acceleration. It is a result of algebraic manipulation and the fact that the gravitational force near the Earth's surface can be approximated as constant. This is due to the small correction terms from the Taylor Series expansion being negligible at this scale. Therefore, the equation is a simplified version of the actual force equation and is commonly used in high school physics.
  • #1
member 428835
hey all!

can anyone please explain where the equation [itex]{Gmm_e}/{r^2}={mgR^2/r^2}[/itex] comes from given $$ma=Gmm_e/r^2$$ where [itex]G,m,m_e[/itex] are the universal gravity constant, mass of object, Earth's mass and [itex]r[/itex] is the distance from center of Earth's mass and [itex]R[/itex] is radius of earth

thanks!
 
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  • #2
joshmccraney said:
hey all!

can anyone please explain where the equation [itex]{Gmm_e}/{r^2}={mgR^2/r^2}[/itex] comes from given $$ma=Gmm_e/r^2$$ where [itex]G,m,m_e[/itex] are the universal gravity constant, mass of object, Earth's mass and [itex]r[/itex] is the distance from center of Earth's mass and [itex]R[/itex] is radius of earth

thanks!


Hi.

[itex]{Gmm_e}/{r^2}={mgR^2/r^2}[/itex] means [itex]\frac{Gm_e}{R^2}=g[/itex]
g is acccelation by Earth on the surface.
 
  • #3
sweet springs said:
Hi.

[itex]{Gmm_e}/{r^2}={mgR^2/r^2}[/itex] means [itex]\frac{Gm_e}{R^2}=g[/itex]
g is acccelation by Earth on the surface.

yes this is true from algebraic manipulation but where does the identity come from?
 
  • #4
Hi.

Acceleration on Earth's surface is given by g.
Force acting on mass m is mg by Newton's law.
 
  • #5
sweet springs said:
Hi.

Acceleration on Earth's surface is given by g.
Force acting on mass m is mg by Newton's law.

yes, i understand this but where does the initial equality derive from?
 
  • #6
[itex]
F=\frac{GM_e m}{r^2} \\
g=\frac{GM_e}{R_e^2} \\
F=\frac{GM_e}{R_e^2} \frac{m R_e^2}{r^2} \Rightarrow F=\frac{gmR_e^2}{r^2}
[/itex]
 
  • #7
According to Newton's 2nd law we know that when the force of magnitude F act's upon an object of mass m it causes a body to accelerate at a. F=ma, in case of the gravitational force with is [itex]F=G\frac{mM}{r^2}[/itex] it is usual to denote this acceleration g so:

[itex]F=mg=G\frac{mM}{r^2}[/itex], dividing by mass gives your expression.
 
  • #8
joshmccraney said:
yes this is true from algebraic manipulation but where does the identity come from?
eh? we start by defining
[tex]\frac{Gm_e}{R^2}=g[/tex]
Then we get the identity.
 
  • #9
joshmccraney, the equality comes from the fact that close to the Earth's surface the gravitational force can be regarded as being approximately constant, i.e., as not depending on distance as measured from the ground.

It is easy to see this if you expand the gravitational force in a Taylor Series around [itex]R_{E}[/itex], where [itex]R_{E}[/itex] is the radius of the Earth. If you do that, you will get the following series:

[itex]F(R_{E} + r ) = \frac{GM_e m}{( R_{E} + r )^2} = \frac{GM_e m}{R_{E}^2} + \frac{-2}{1!}\frac{GM_e m}{R_{E}^3}r + \frac{6}{2!}\frac{GM_e m}{R_{E}^4}r^2 + \centerdot\centerdot\centerdot[/itex]

So this is an expression for the graviational force close to the Earth's surface, being [itex]r[/itex] just the distance or height you measure from the ground.

Ok, we now we have this infinite series, which looks kind of weird because we wouldn't like to calculate infinite terms to get the actual value of the force near the planet's surface. But, take a closer look at the terms of the expansion and you will see that all of them can be written as the first term multiplied by a constant times [itex]( \frac{r}{R_{E}} )^n[/itex], with [itex]n ≥ 1[/itex] being an integer. How big is that number? We know that [itex]R_{E} ≈ 6.4×10^6 m[/itex], and say [itex]r ≈ 100m[/itex]. If you plug these values into the expression you get:

[itex]( \frac{r}{R_{E}} )^n = ( \frac{100}{6.4×10^6} )^n ≈ 10^{-5n}[/itex]

This means that the second term of the expansion (for which [itex]n=1[/itex]) will make a correction to the fisrt term on the fifth decimal place - it's a very small correction! For the third term ( [itex]n = 2[/itex] ), for instance, the correction is even smaller since it would take place on the tenth decimal place! We conclude then that all these terms (from the second on) are too small for our purposes, since we are interested on the graviational force on the surface's neighborhood. We can then discard these terms and keep just the first one:

[itex]F(R_{E} + r ) = \frac{GM_e m}{R_{E}^2}[/itex]

Now, because we don't like to write all these constants, we simply define [itex]g = \frac{GM_e}{R_{E}^2}[/itex] and get the famous expression we all learn in high school:

[itex]F = mg[/itex]Zag
 
Last edited:

What is gravity?

Gravity is a natural force that causes objects with mass to attract each other. It is responsible for keeping planets in orbit around the sun and for objects falling to the ground.

How does gravity work?

Gravity is caused by the curvature of spacetime around objects with mass. This curvature creates a force that pulls objects towards each other.

Why do objects fall to the ground?

Objects fall to the ground because the mass of the Earth creates a gravitational force that pulls objects towards its center. This force is stronger on objects with more mass, which is why heavier objects fall faster than lighter objects.

What is the difference between weight and mass?

Mass is a measure of the amount of matter in an object, while weight is a measure of the force of gravity acting on an object. Mass is constant, while weight can vary depending on the strength of gravity.

How does gravity affect the motion of objects?

Gravity affects the motion of objects by pulling them towards the center of the Earth. This force can change the speed and direction of an object's motion, causing it to accelerate towards the ground.

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