A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 7.00×10^-3 m/s^2 less than that at sea level. What is the observatory's altitude?

Answer 1

From newtons law:

force=GMe*M/r^2
but force/mass= acceleration so we get

g=GMe/r^2 and when r= radiusEarth, g works out to be 9.81m/s^2

so where does g = 9.81-.007 m/s^2 ?

GMe/re^2-GMe/(re+h)^2 = .007
there is a bit of algebra here to solve for h, I admit, but you can do that.

I have a bumper sticker: Girl Scouts can do Anything. I assume you were a girl scout.

I can check your work if you need.

YOu might want to do this: divide both sides by re^2, that gives...

GMe-GMe/(1+h/re)^2 = .007/re
then change that to..
GMe-GMe/(1+x)^2 = .007/re

then solve for x, when you get it, then solve for h.
Have fun.

Answer 2

oops, the right side should be .007/re^2
even GScout leaders make mistakes.

Answer 3

When you divided by re^2, shouldn't you have multiplied by re^2?

Answer 4

Actually, to do it the way he did would be wrong all together, because you would have to foil the (1+h/re)^2... you would have to multiply bothe sides by (1+h/re)^2 first... giving you:

GMe-GMe=.007(1+h/re)^2

Obviously, GMe-GMe would give you zero... creating a quadractic equation once you reverse foil the thing... algebra works like this:

0=.007(1+h/re)^2

let x=h/re

0=.007(1+x)^2

this equals:

0=.007(1+x)(1+x)

0=.007(1+2x+x^2)

0=.007x^2+.014x+.007

once you solve for x, you would set x back = to h/re and solve for h...

This is assuming that he also did the rest of his algebra correct... I haven't checked that yet...

Answer 5

which it's not at all... if you start with g=GMe/re^2 then the difference in the two gravitys would be:

(GMe/re^2)-(GMe/(h-re)^2=.007

to get like sides here, start by factoring out a GME... leaving you with

GMe(1/re^2 - 1/(h-re)^2)=.007

divide both sides by GMe and you have:

(1/re^2)-(1/(h-re)^2)=.007/GMe

now, this gets tricky, so follow closely. if you divide both sides by (1/re^2)-(1/(h-re)^2) you get:

1=(.007/(GMe*(1/re^2-(1/(h-re)^2))

that goes to 1=.007/((GMe/re^2)-(GMe/(h-re)^2)

which expands to:

1=.007/(GMe/re^2) - .007/(GMe/(h-re)^2

if you divide something, you are actually multiplying it by it's reciprical so you have:

1=(.007re^2/GMe)-(.007(h-re)^2/GMe)

multiply out GMe to both sides:

GMe=.007re^2-.007(h-re)^2

that expands out to

GMe=.007re^2-.007(h^2-hre+re^2)

divide out .007:

GMe/.007=re^2-h^2-hre+re^2

this simplifies to

GMe/.007=-h^2-hre+2re^2

subtract both sides by -h^2-hre+2re^2 to be able to set up your quadratic equation like above....

so you are left with:

h^2+hre-2re^2+GMe/.007=0

your quadratic equation would be

h= -re+/-sqrt((re^2-(4*1*(-2re^2+GMe/.007)))/2*1)

plug in your GMe and re and pick which ever one makes sense and you have your h values :)

Answer 6

WAIT!!!!! i did h-r instead of h+r... therefore rework the steps with h+re instead of h-re..... sorry