# Simple gravity question

by joshmccraney
Tags: gravity, simple
 P: 346 hey all! can anyone please explain where the equation ${Gmm_e}/{r^2}={mgR^2/r^2}$ comes from given $$ma=Gmm_e/r^2$$ where $G,m,m_e$ are the universal gravity constant, mass of object, earth's mass and $r$ is the distance from center of earths mass and $R$ is radius of earth thanks!
P: 449
 Quote by joshmccraney hey all! can anyone please explain where the equation ${Gmm_e}/{r^2}={mgR^2/r^2}$ comes from given $$ma=Gmm_e/r^2$$ where $G,m,m_e$ are the universal gravity constant, mass of object, earth's mass and $r$ is the distance from center of earths mass and $R$ is radius of earth thanks!

Hi.

${Gmm_e}/{r^2}={mgR^2/r^2}$ means $\frac{Gm_e}{R^2}=g$
g is acccelation by earth on the surface.
P: 346
 Quote by sweet springs Hi. ${Gmm_e}/{r^2}={mgR^2/r^2}$ means $\frac{Gm_e}{R^2}=g$ g is acccelation by earth on the surface.
yes this is true from algebraic manipulation but where does the identity come from?

 P: 449 Simple gravity question Hi. Acceleration on earth's surface is given by g. Force acting on mass m is mg by Newton's law.
P: 346
 Quote by sweet springs 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???
 P: 868 $F=\frac{GM_e m}{r^2} \\ g=\frac{GM_e}{R_e^2} \\ F=\frac{GM_e}{R_e^2} \frac{m R_e^2}{r^2} \Rightarrow F=\frac{gmR_e^2}{r^2}$
 P: 65 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 $F=G\frac{mM}{r^2}$ it is usual to denote this acceleration g so: $F=mg=G\frac{mM}{r^2}$, dividing by mass gives your expression.
HW Helper
P: 3,448
 Quote by joshmccraney yes this is true from algebraic manipulation but where does the identity come from?
eh? we start by defining
$$\frac{Gm_e}{R^2}=g$$
Then we get the identity.
 P: 43 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 $R_{E}$, where $R_{E}$ is the radius of the Earth. If you do that, you will get the following series: $F(R_{E} + r ) = \frac{GM_e m}{( R_{E} + r )^2} = \frac{GM_e m}{R_{E}^2} + \frac{-2}{1!}\frac{GM_e m}{R_{E}^3}r + \frac{6}{2!}\frac{GM_e m}{R_{E}^4}r^2 + \centerdot\centerdot\centerdot$ So this is an expression for the graviational force close to the Earth's surface, being $r$ 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 $( \frac{r}{R_{E}} )^n$, with $n ≥ 1$ being an integer. How big is that number? We know that $R_{E} ≈ 6.4×10^6 m$, and say $r ≈ 100m$. If you plug these values into the expression you get: $( \frac{r}{R_{E}} )^n = ( \frac{100}{6.4×10^6} )^n ≈ 10^{-5n}$ This means that the second term of the expansion (for which $n=1$) will make a correction to the fisrt term on the fifth decimal place - it's a very small correction! For the third term ( $n = 2$ ), 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: $F(R_{E} + r ) = \frac{GM_e m}{R_{E}^2}$ Now, because we don't like to write all these constants, we simply define $g = \frac{GM_e}{R_{E}^2}$ and get the famous expression we all learn in high school: $F = mg$ Best, Zag

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