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?
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.
oops, the right side should be .007/re^2
even GScout leaders make mistakes.
When you divided by re^2, shouldn't you have multiplied by re^2?
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...
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 :)
WAIT!!!!! i did h-r instead of h+r... therefore rework the steps with h+re instead of h-re..... sorry
To determine the observatory's altitude given the difference in free-fall acceleration, we can use the following equation:
Δg = g_sea - g_obs
where:
Δg = Difference in free-fall acceleration
g_sea = Free-fall acceleration at sea level (9.8 m/s^2)
g_obs = Free-fall acceleration at the observatory
First, let's rearrange the equation to solve for g_obs:
g_obs = g_sea - Δg
Now, substitute the known values into the equation:
g_obs = 9.8 m/s^2 - 7.00×10^-3 m/s^2
Calculate the subtraction:
g_obs = 9.8 m/s^2 - 0.007 m/s^2
g_obs = 9.793 m/s^2
Now that we have the free-fall acceleration at the observatory, we can use this value to determine the altitude of the observatory. The acceleration due to gravity decreases with an increase in altitude, so we can use this relationship to find the observatory's height. We can express this relationship using the formula for gravitational acceleration:
g = G * (M / r^2)
where:
g = Free-fall acceleration
G = Universal gravitational constant (6.674 × 10^-11 N m^2/kg^2)
M = Mass of the Earth (5.972 × 10^24 kg)
r = Distance from the center of the Earth to the observatory (radius)
Since we are interested in finding the altitude of the observatory, we need to determine the distance from the center of the Earth to the observatory (r). This can be calculated using the formula:
r = R + h
where:
R = Radius of the Earth (6.371 × 10^6 m)
h = Height or altitude of the observatory
By substituting the known values into the equation, we can solve for h:
r = (6.371 × 10^6 m) + h
Now we can substitute the value of r into the gravitational acceleration formula:
g_obs = G * (M / (6.371 × 10^6 m + h)^2)
Let's rearrange the equation to solve for h:
h = (g_obs * (6.371 × 10^6 m + h)^2) / G
Although a direct calculation of h might be complicated due to the squared term, we can use an iterative process to approximate its value. Here's a step-by-step approach:
1. Start with an initial guess for h, e.g., 10,000 meters.
2. Substitute this guess into the equation and calculate a new value for h.
3. Repeat step 2 until the value of h stops changing significantly with each iteration.
4. The final value of h will be our estimated altitude of the observatory.
Keep in mind that this approximation assumes a spherically symmetric Earth and neglects other factors that may impact gravitational acceleration, such as local variations in the Earth's density.