The weight of an object of mass m and altitude x miles above the surface of the earth is: w(x)=mgR^2/(R+x)^2.?

Question

(where R is the radius of the Earth and g is the acceleration due to gravity)

Show that w(x) = mg(1-2x/R). Estimate how large x would need to be to reduce the weight by 10%.

Answer 1

w(x) = mgR^2/(R+x)^2 = mgR^2/(R^2+2Rx+x^2)
Since R>>x, we can neglect x^2 in the denominator.
w(x)
= mgR^2/(R^2+2Rx)
= mg/(1+2x/R)
= mg[1-2x/R+(2x/R)^2-...]
= mg[1-2x/R], neglected higher degree terms of x/R.

mg[1-2x/R] = (90%) mg
Solve for x,
x = 0.05R, estimated value of x.

Answer 2

w(x) = mgR^2/(R + x)^2
w(x) = mg(R/(R + x))^2
w(x) = mg(1/(1 + x/R))^2
w(x) = mg (1/(1 + 2x/R + x^2/R^2)
let a = x/R
.. . . . . . . . . . 1 - 2a + 3a^2 - 4a^3
1 + 2a + a^2)1
. . . . . . . . . . 1 + 2a + a^2
. . . . . . . . . . . . . - 2a - a^2
. . . . . . . . . . . . . - 2a - 4a^2 - 2a^3
. . . . . . . . . . . . . . . . + 3a^2 + 2a^3
. . . . . . . . . . . . . . . . + 3a^2 + 6a^3 + 3a^4
. . . . . . . . . . . . . . . . . . . . . . . - 4a^3 - 3a^4
w(x) ≠ mg(1-2x/R)
w(x) ≈ mg(1 - 2x/R + 3(x/R)^2 - 4(x/R)^3 + . . . . . . + n(-x/R)^(n - 1))
w(x) ≈ mg(1 - 2x/R) holds fairly well if x < R/100
Using
w(x) ≈ mg(1 - 2x/R),
mg(1 - 2x/R) ≈ 0.9mg
2x/R ≈ 0.1
x ≈ 0.05R

Using
mgR^2/(R + x)^2
mgR^2/(R + x)^2 = 0.9mg
R^2 = 0.9(R^2 + 2xR + x^2)
0.9x^2 + 1.8xR - 0.1R^2 = 0
9x^4 + 18xR - R^2 = 0
x = R(-18 ± √(324 + 36))/18
x = R(-1 ± √(1 + 1/9)
x = R(-1 ± (1/3)√(10)
x ≈ 0.054093R
so the approximation using only the 1st expansion term is in error by 7.6%