Recent content by Dietrichw

yes, if q is really small it would be approximately 1. I don't understand the 1+q part.

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}

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...

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

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

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})

\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?

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

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

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.

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 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 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

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

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