# Law of universal gravitation and Kepler's law

• Dietrichw
Better still, use part 1. what is the change in a function whose derivative is constant?I don't fully understand what you are saying.f

## Homework Statement

Show that the rate of change of the free-fall acceleration with vertical position near the Earth's surface is
$$\frac{d}{dr} [g] = -\frac{2GM_E}{R_E^3}$$

Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is
$$\Delta g = \frac{(2GM_E h)}{R_E^3}$$

Show that the minimum period for a satellite in orbit around a spherical planet of uniform density ρ is
$$T_{min} = \sqrt{ \frac{3\pi}{Gp}}$$

## Homework Equations

$$g=\frac{GM_E}{R_E^2}$$
$$T^2=Kr^3$$

## The Attempt at a Solution

I did the first one easily
$$\frac{d}{dr} [\frac{GM_E}{R_E^2}]$$
$$\frac{d}{dr} [GM_E R_E^{-2}]$$
simple power rule
The second and third not so much, I substituted one R with (R+h) as that is what it seems like is going on and find the difference. That did not work out for me
The third I thought about p= Mass/Volume and solving for R and substitute but I'm not sure what to do with K and Kepler's 3rd observation is the only equation that I have the deals with the period

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For part 1, ##R_E## is constant so you can't differentiate your formula with respect to ##r##.

For part 1, ##R_E## is constant so you can't differentiate your formula with respect to ##r##.
yes, in the situation described using Earth it would be a constant. I think they only did that to make it relatable but still treat it as a variable as it could be used for other bodies in space

yes, in the situation described using Earth it would be a constant. I think they only did that to make it relatable but still treat it as a variable as it could be used for other bodies in space
You could differentiate the formula for ##r## then replace ##r## with ##R_E##. That would be valid. What you have is not valid.

What went wrong for part 2?

You could differentiate the formula for ##r## then replace ##r## with ##R_E##. That would be valid. What you have is not valid.

What went wrong for part 2?
when trying to simplify I get a bigger mess that I don't know what to do with

$$\frac{-2GM_E}{R_E^3} - \frac{-2GM_E}{(R_E+h)^3}$$

expanding (R+h)^3 gets me $$R^3 +3R^2h+3Rh^2+h^3$$
simplifying from there makes it worse

when trying to simplify I get a bigger mess that I don't know what to do with

$$\frac{-2GM_E}{R_E^3} - \frac{-2GM_E}{(R_E+h)^3}$$

expanding (R+h)^3 gets me $$R^3 +3R^2h+3Rh^2+h^3$$
simplifying from there makes it worse
That's ##\Delta(dg/dr)## you are trying to calculate there.

The point of part was to show that ##dg/dr## is approximately constant.

It's ##\Delta g## you need to calculate.

That's ##\Delta(dg/dr)## you are trying to calculate there.

The point of part was to show that ##dg/dr## is approximately constant.

It's ##\Delta g## you need to calculate.
I thought so, I tried it with $$\frac{GM_E}{R_E^2}$$ but ran into the same issue at the same point, so I tried both

I thought so, I tried it with $$\frac{GM_E}{R_E^2}$$ but ran into the same issue at the same point, so I tried both
That should have worked. You need to use the binomial expansion doing it that way.

Instead, can you see how to use part 1? It's very simple. Compare the answers to parts 1 and 2.

That should have worked. You need to use the binomial expansion doing it that way.

Instead, can you see how to use part 1? It's very simple. Compare the answers to parts 1 and 2.
I expanded (R+h)^2 to
$$R^2+2Rh+h^2$$
multiplied to get a common denominator
$$\frac {(R^2 +2Rh+h^2)GM-R^2GM}{R^2(R^2 +2Rh+h^2)}$$

I could only remove R^2GM after that

I expanded (R+h)^2 to
$$R^2+2Rh+h^2$$
multiplied to get a common denominator
$$\frac {(R^2 +2Rh+h^2)GM-R^2GM}{R^2(R^2 +2Rh+h^2)}$$

I could only remove R^2GM after that

The point of part 1 was to avoid this. You don't need to expand ##(R+h)^2##. Instead take out a factor of ##R^2## and use the binomial expansion.

Better still, use part 1. what is the change in a function whose derivative is constant?

The point of part 1 was to avoid this. You don't need to expand ##(R+h)^2##. Instead take out a factor of ##R^2## and use the binomial expansion.

Better still, use part 1. what is the change in a function whose derivative is constant?
I don't fully understand what you are saying.
What I am getting is
$$\frac{GM}{R^2} - \frac {GM}{(R+h)^2}$$
then factor a ##R^2## from each term?

I don't fully understand what you are saying.
What I am getting is
$$\frac{GM}{R^2} - \frac {GM}{(R+h)^2}$$
then factor a ##R^2## from each term?
yes, why not?

yes, why not?
so...

$$\frac{1}{R^2}(GM- \frac{GMR^2}{(R+h)^2} )$$

I haven't done anything like this so I'm not too confident in what I am doing.

so...

$$\frac{1}{R^2}(GM- \frac{GMR^2}{(R+h)^2} )$$

I haven't done anything like this so I'm not too confident in what I am doing.

##\frac{GM}{R^2}(1 - \frac{1}{(1 +h/R)^2})##

##\frac{GM}{R^2}(1 - \frac{1}{(1 +h/R)^2})##
oh yea, I could have also taken out ##GM## as well.
The ##\frac{1}{(1+h/R)^2}## is really confusing me

oh yea, I could have also taken out ##GM## as well.
The ##\frac{1}{(1+h/R)^2}## is really confusing me

##R^2(1+h/R)^2 = (R+h)^2##

##R^2(1+h/R)^2 = (R+h)^2##
okay so
$$R^2(1+h/R)^2 = R(1+h/R)R(1+h/R) = (R+hR/R)^2= (R+h)^2$$
that is really cool how that works out

The point of part 1 was to avoid this. You don't need to expand ##(R+h)^2##. Instead take out a factor of ##R^2## and use the binomial expansion.

Better still, use part 1. what is the change in a function whose derivative is constant?
$$\frac{GM}{R^2}(1-\frac{1}{(1+h/R)^2})$$ can be rewritten as
$$\frac{GM}{R^2}(\frac{(1+h/R)^2-1}{(1+h/R)^2})$$

Then do the "binomial expansion" as you said?

$$\frac{GM}{R^2}(1-\frac{1}{(1+h/R)^2})$$

simply apply the binomial expansion to ##(1 + h/R)^{-2}##

simply apply the binomial expansion to ##(1 + h/R)^{-2}##
so
$$(1+h/R)^{-1} (1+h/R)^{-1} = (1+2(h/R) +h^2/R^2)^{-1}$$

$$\frac{GM}{R^2} (1-(1+2(h/R) +h^2/R^2)^{-1})$$

so
$$(1+h/R)^{-1} (1+h/R)^{-1} = (1+2(h/R) +h^2/R^2)^{-1}$$

$$\frac{GM}{R^2} (1-(1+2(h/R) +h^2/R^2)^{-1})$$
Thats not right at all. you need to brush up your maths. I'm going off line now.

Thats not right at all. you need to brush up your maths. I'm going off line now.
okay.

even dumming it down to (a+b)^2 = 1a+2ab+b^2
a=1 b= h/R so ## 1^2 +2*(1)(h/R)+ (h/R)^2##

something must be going right over my head if I am missing something so simple

I'm still not getting what is wrong here. Am I not getting what you mean by binomial expansion?

I don't fully understand what you are saying.
What I am getting is
$$\frac{GM}{R^2} - \frac {GM}{(R+h)^2}$$
then factor a ##R^2## from each term?
Yes, factor out GM/R2 from each term.
You have problem with the second term then, ##\left ( \frac{1}{1+\frac{h}{R}} \right )^2 ##
(h/R) is small, denote q=-h/R. So you have the expression ##\frac{1}{1-q}##. That should remind you on something: the sum of infinite geometric series.If you have a geometric progression 1, q, q2, q3...qn, what is the sum of the terms? What is the limit of the sum when n goes to infinity?

Yes, factor out GM/R2 from each term.
You have problem with the second term then, ##\left ( \frac{1}{1+\frac{h}{R}} \right )^2 ##
(h/R) is small, denote q=-h/R. So you have the expression ##\frac{1}{1-q}##. That should remind you on something: the sum of infinite geometric series.If you have a geometric progression 1, q, q2, q3...qn, what is the sum of the terms? What is the limit of the sum when n goes to infinity?
I see the resemblance to the geometric series as ##h/R## would be less than one and would converge if it was a ratio in a geometric series but I don't see how that relates to the equation.
As a side note. My progression through calculus is that I am taking a test on diverging and converging infinite series and Taylor polynomials tomorrow morning.

I see the resemblance to the geometric series as ##h/R## would be less than one and would converge if it was a ratio in a geometric series but I don't see how that relates to the equation.
As a side note. My progression through calculus is that I am taking a test on diverging and converging infinite series and Taylor polynomials tomorrow morning.
OK, so you have learned how the derivative of a function is defined?
And what is the sum of the geometric series 1+q+q2+...qn?

OK, so you have learned how the derivative of a function is defined?
And what is the sum of the geometric series 1+q+q2+...qn?

Not exactly sure how to define it but I understand what subsequent derivatives of a function represent tangentcy and concavity etc. Not something I think about normally

if q is less than 1 it would converge to $$\frac {1}{1-q}$$

if q is less than 1 it would converge to $$\frac {1}{1-q}$$
Well, so you can write that 1+q+q2+q3+...=1/(1-q) .
If q<<1 , you can ignore all higher powers of q and approximate 1/(1-q) by 1+q. Note it,
##\frac{1}{1-q}≈1+q##.
It is a very useful approximation.
In your problem, q = -h/R. You can use this approximation in the expression
##
\frac{GM}{R^2} - \frac {GM}{(R+h)^2}=\frac{GM}{R^2}(1-(\frac{1}{1+h/R})^2)## .

• Vibhor
Well, so you can write that 1+q+q2+q3+...=1/(1-q) .
If q<<1 , you can ignore all higher powers of q and approximate 1/(1-q) by 1+q. Note it,
##\frac{1}{1-q}≈1+q##.
It is a very useful approximation.
In your problem, q = -h/R. You can use this approximation in the expression
##
\frac{GM}{R^2} - \frac {GM}{(R+h)^2}=\frac{GM}{R^2}(1-(\frac{1}{1+h/R})^2)## .
yes, if q is really small it would be approximately 1. I don't understand the 1+q part.

yes, if q is really small it would be approximately 1. I don't understand the 1+q part.
q is not so small, that it can be fully ignored. You need the difference ##(1-(\frac{1}{1+h/R})^2)##. If you ignore h/R, you get zero for the difference between g-s at two points which are close. But it is not true that g is the same!
So keep the first power of q and ignore the others from the sum 1+q+q2+q3+...
Just to see how it works, calculate 1/(1-0.0001)-1 with 4 significant digits.

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• Vibhor