Geometric Progressions: Finding the Sum of an Infinite Series

[faiziqb12]

using Newtons equation for gravitational force
one can fling that the acceleration due to gravity, g' above the surface of Earth as

g' = g (d/(d+h))

but I find that there is a different equation for the same as

g' = g ( 1- 2h/r )

I know the first formula is right
but I can't doubt the first formula...
however the unit of h and r in the second formula is km as compared to the m of the first formula
please help me derive the second formula

 

[Chestermiller]

The first formula is wrong, and the second formula is an approximation to the corrected version of the first formula for small values of h/r.

Chet

 

[faiziqb12]

The first formula is wrong, and the second formula is an approximation to the corrected version of the first formula for small values of h/r.

Chet

so as you have said the first formula is wrong
then please show how its wrong

and yes please show the derivation method of the second equation

thanks

 

[Chestermiller]

so as you have said the first formula is wrong
then please show how its wrong

and yes please show the derivation method of the second equation

thanks

Gravity varies inversely with distance (a) to the first power or (b) to the second power?

 

[faiziqb12]

Gravity varies inversely with distance (a) to the first power or (b) to the second power?

obviously to the 2nd power

 

[faiziqb12]

The first formula is wrong, and the second formula is an approximation to the corrected version of the first formula for small values of h/r.

Chet

I'm sorry but I missed the square in my second formula

the correct one is
g' = g (d / ( d+h )) ^ 2

 

[Chestermiller]

I'm sorry but I missed the square in my second formula

the correct one is
$g' = g (d / ( d+h )) ^ 2$

Good. Now let x = h/d. Please re-express your equation in terms of x.

 

[faiziqb12]

Good. Now let x = h/d. Please re-express your equation in terms of x.

the new formula is then
g' = g ( 1 / (1+x) ) ^ 2

 

[faiziqb12]

what am I supposed to do further
please suggest

 

[Chestermiller]

the new formula is then
g' = g ( 1 / (1+x) ) ^ 2

Good, this is correct even with all those parentheses. But, here's a piece of advice: if you don't simplify your mathematical expressions when you are at your present stage, you are going to encounter real problems (manipulating the mathematics) when you get to more complicated analyses. So, I'm going to simplify it for you:
$$g'=\frac{g}{(1+x)^2}$$
Have you gotten far enough in calculus to be able to expand this in a Taylor series about x = 0?

Chet

 

[faiziqb12]

Good, this is correct even with all those parentheses. But, here's a piece of advice: if you don't simplify your mathematical expressions when you are at your present stage, you are going to encounter real problems (manipulating the mathematics) when you get to more complicated analyses. So, I'm going to simplify it for you:
$$g'=\frac{g}{(1+x)^2}$$
Have you gotten far enough in calculus to be able to expand this in a Taylor series about x = 0?

Chet

I'm sorry
but I can't. I'm just studying in class 9th
isn't there an alternative method

 

[Chestermiller]

I'm sorry
but I can't. I'm just studying in class 9th
isn't there an alternative method

Yes. From what you learned about geometric progressions, what is the infinite sum 1-x+x²-x³... equal to?

 

[jtbell]

Or look up the "binomial approximation." It's a very useful thing to know for situations like this.

 

[faiziqb12]

Yes. From what you learned about geometric progressions, what is the infinite sum 1-x+x²-x³... equal to?

thanks chestmiller
looks like I can do it now

it would 've your goodness if you show me the whole method involving this progression