ghostbusters911
We treated the gravitational field of the earth as a constant (9.80 m/s2), but we know that the gravitational field of the earth should fall off as 1/r2. How far away from the surface of the earth would we have to go for the field to decrease in the last decimal place, which is the 0.1% level? In other words, over what distance from the surface of the earth is field essentially constant to the 0.1% level? Treat the earth a uniform sphere of mass with radius 6370 km.

attempt:
used the equation g= Gm/r^2

mutiplied 9.8(.999) to get 9.7902 = g
G=6.67e-11
Mass earth= 5.97e24

i plugged values into equation g=Gm/r^2 to find r.
Then i subtracted r from radius of earth to get 9799.25m (distance away from surface of earth that has 9.7902 gravitational pull.)

ive tried multiplying 9.8 by .001 and tried doing that way to, but to no avail..maybe im just not understanding the question?

rpenner
You are doing it in an imprecise way because
( 6.67e-11 × 5.97e24) / (6370000)^2 = 9.81344+
not 9.80.

You get more precise numbers if you just use R, the height and the fractional change in g.

Starting with:
g = GM/R^2
x g = GM/(R+y)^2

I can substitute the right side of the first equation for g in the second equation.
x GM/R^2 = GM/(R+y)^2
So
(R+y)^2 = R^2/x
R + y = R/sqrt(x)
y = R/sqrt(x) - R
y = (1/sqrt(x) - 1)R

The phrasing is awkward, but it looks like they want you to use x = 0.999

So I get about 3.2 km.
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