Showing that a near earth orbit is P = C(1 + 3h/2R_E)

[jeanbeanie]

Homework Statement

Show that a satellite in low-Earth orbit is approximately P = C(1 + 3h/2R_E) where h is the height of the satellite, C is a constant, and R_E is the radius of the earth)

Homework Equations

p=C(1+3H/2R_E)
p² = (4∏²)/(GM) * a³. (G - gravitational constant, M - mass of the Earth (in this case) and a = semi-major axis)

The Attempt at a Solution

I've been googling this and I seen one person say use Taylor series. We have not touched that so I am pretty sure it isn't how I should be finding this

so the a is equal to the radui of earth, and its high
a=h+R_E
and i can write that as

a=R_E(1+h/R_E)

SO p² = (4∏²)/(GM) * R_E(1+h/R_E)³.

I just saw in my noes that's if (4∏²)/(GM) can be cancled out if all meaurments are in Au's, years and solars mass.
Then Id get
p² = R_E(1+h/R_E)³
and p= R_E(1+h/R_E)^3/2

Now what?

 

[voko]

Now you have to recall that $\frac {df(x) } {dx} = \lim_{\Delta x \rightarrow 0} \frac {f(x + \Delta x) - f(x)} {\Delta x}$ which for some small $\Delta x$ means that $f(x + \Delta x) - f(x) \approx \frac {df(x) } {dx} \Delta x$.

 

[jeanbeanie]

Now you have to recall that $\frac {df(x) } {dx} = \lim_{\Delta x \rightarrow 0} \frac {f(x + \Delta x) - f(x)} {\Delta x}$ which for some small $\Delta x$ means that $f(x + \Delta x) - f(x) \approx \frac {df(x) } {dx} \Delta x$.

we have never done anything like this, I am not even sure were to start with it

 

[voko]

we have never done anything like this, I am not even sure were to start with it

You do not know what the derivative of a function is?

 

[RockenNS42]

You do not know what the derivative of a function is?

we we have done that, but I didnt get the second part
but now that I am more awake and looking at it again, does it mean
just take the dertive of the function that I already had?

 

[voko]

The second part means that if you know the value of a function and its derivative at some point x, then you can approximate the value of the function at some close point, via the expression given. Note that you have a function which you want to compute at (1 + h/R). The function (and the derivative) at 1 are trivial, and (1 + h/R) is very close to 1 because h is much smaller than R.

 

[RockenNS42]

The second part means that if you know the value of a function and its derivative at some point x, then you can approximate the value of the function at some close point, via the expression given. Note that you have a function which you want to compute at (1 + h/R). The function (and the derivative) at 1 are trivial, and (1 + h/R) is very close to 1 because h is much smaller than R.

So do you mean that p= R_E(1+h/R_E)^3/2 is approximately
p= R_E(1*^3/2
then
p= R_E
?
Then the derivative would be p=0
Or am I really wrong?

 

[voko]

As shown above, $f(x + \Delta x) - f(x) \approx \frac {df(x) } {dx} \Delta x$. What would $x$ and $\Delta x$ be in this case?

 

[RockenNS42]

As shown above, $f(x + \Delta x) - f(x) \approx \frac {df(x) } {dx} \Delta x$. What would $x$ and $\Delta x$ be in this case?

isnt x the function?
p= R_E(1+h/R_E)^3/2

and found Δx be the h/R because it is going to zero?

 

[voko]

x is the independent variable. What is the function in your problem?

 

[RockenNS42]

x is the independent variable. What is the function in your problem?

Ok, I really thought that the p= R_E(1+h/R_E)^3/2 was the function
as in
p(h)= R_E(1+h/R_E)^3/2

 

[voko]

That is one possible choice of the function. Now you need to choose x and consequently Δx so that h = x + Δx. These are usually chosen so that f(x) and f'(x) are very simple.

 

[RockenNS42]

That is one possible choice of the function. Now you need to choose x and consequently Δx so that h = x + Δx. These are usually chosen so that f(x) and f'(x) are very simple.

like any x? like x and 1?
Or something to do with the problem like
R_E and 0?

Or would it have nothing do to with the eq?

Im really sorry, we never went over any of this is in class

 

[voko]

Consider this very simple example. f(x) = x^2. Compute f(1.000001) using the derivative (the formula above).

 

[RockenNS42]

Consider this very simple example. f(x) = x^2. Compute f(1.000001) using the derivative (the formula above).

right ok, woudlnt I just plug in the value for x as 1.000001?
Like

f(1.000001) = 1.000001^2
and
f^'(x)=2x
f^'(1.000001)=2(1.000001)

 

[voko]

That did not use the f'(x)Δx formula.

 

[RockenNS42]

That did not use the f'(x)Δx formula.

ok

f(x+Δx)−f(x)≈df(x)/dx Δx

so 2x*Δx ? I do not understand what the Δx even is in this case

 

[voko]

Do you see that if x = 1, then both f(x) and f'(x) are very simple? What is then Δx and what is f(x + Δx)?

 

[RockenNS42]

Do you see that if x = 1, then both f(x) and f'(x) are very simple? What is then Δx and what is f(x + Δx)?

I do see that f(x) = x^2 it would be 1
and in f'(x) it would be 2


f(x+Δx)−f(x)≈df(x)/dx Δx
f(x+Δx)−1≈2 Δx
f(x+Δx)≈2 Δx +1
Like I don't understand how to use the formula maybe? I understand what a derive is and how to plug and chug numbers, just not this forumla or even how it applies to my problem. I am sorry

 

[voko]

f(x+Δx)≈2 Δx +1 is not the complete answer. You were requested to compute f(1.000001) using the formula. You have the formula, but not the numeric result.

 

[RockenNS42]

does that mean that x+Δx is the 1.000001?
Making Δx (1.000001 -1)/2
?

 

[voko]

does that mean that x+Δx is the 1.000001?

Yes. x + Δx must always be equal to whatever value we want to compute the function at.

Making Δx (1.000001 -1)/2

What does this mean?

 

[RockenNS42]

Yes. x + Δx must always be equal to whatever value we want to compute the function at.



What does this mean?

i ment Δx = (1.000001 -1)/2 which is really small? getting close to zero...

 

[voko]

If Δx = (1.000001 - 1)/2, then x + Δx = 1 + (1.000001 - 1)/2 = 1.0000005, which is not 1.000001.

 

[RockenNS42]

ok so
x + Δx = 1.000001
and i really have no clue.

f(x+Δx)≈2 Δx +1
f(1.000001)≈2 Δx +1? or am i still even using the right thing?

 

[voko]

I do not understand why this is so hard.

x + Δx = 1.000001
x = 1

So 1 + Δx = 1.000001, giving Δx = 0.000001.

Apply the formula and check the result against 1.000001^2.

Then, consider this: f(x) = (1 + x)^2. Find the approximation for very small values of x. Note this function is very similar to the one you have in the original problem.

 

[RockenNS42]

I do not understand why this is so hard.

x + Δx = 1.000001
x = 1

So 1 + Δx = 1.000001, giving Δx = 0.000001.

Apply the formula and check the result against 1.000001^2.

Then, consider this: f(x) = (1 + x)^2. Find the approximation for very small values of x. Note this function is very similar to the one you have in the original problem.

Im sorry that I am having a hard time understanding

so df(x)/dx * Δx the dertivie of x^2 is 2x
so its 2x*1.000001

f(1.000001)≈2 Δx +1? if Δx =0.000001. so 2*0.000001 +1?



f(x) = (1 + x)^2
f'(x)= 2(1+x)

for very small values of x this would be 2, no?

 

[voko]

As I said: Apply the formula and check the result against 1.000001^2.

Since you did not, now compute x^2 using the Δx method for 1.000002 - and please DO check the result against 1.000002^2.

 

[RockenNS42]

the questions been resolved, thanks I went to see the prof