Gravitation Near Earth's Surface — Question 2

Section 13-3 · Physics pages 1934–1935

Question 2 A constant-gravity model needs an explicit error convention A calculation uses g0 at altitude h above a spherical nonrotating Earth. Define its fractional error relative to the true value as (g0 −g(h))/g(h). h/R ϵtrue .01 .00499 True-value-normalized error grows with altitude; the dashed line marks the one-percent threshold (rounded labels). (a) Find the largest h/R for error no greater than 1%. (b) Compare with the result if error were instead normalized by g0.
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Question 2 - Solution A constant-gravity model needs an explicit error convention h/R ϵtrue .01 .00499 (a) Derivation and calculation Set x = h/R ≥0. With true-value normalization, ϵt = g0 −g g = g0 g −1 = (1 + x)2 −1. The one-percent condition yields 2x + x2 ≤.01, x ≤ √ 1.01 −1 = 0.00498756 . (b) Result and interpretation Surface-value normalization instead gives ϵs = g0 −g g0 = 1 −(1 + x)−2 ≤.01. Hence x ≤ 1 √ .99 −1 = 0.00503782 , ϵs = ϵt 1 + ϵt . The larger reference denominator permits a slightly larger altitude. Physical check Both conventions agree to first order, ϵ ≃2h/R, but differ at the precision requested.

Question 2

A constant-gravity model needs an explicit error convention

A calculation uses g0g_0 at altitude hh above a spherical nonrotating Earth. Define its fractional error relative to the true value as (g0−g(h))/g(h)(g_0-g(h))/g(h).

Diagram shown in the original SVG worksheet.

  1. Find the largest h/Rh/R for error no greater than 1%1\%.

  2. Compare with the result if error were instead normalized by g0g_0.

Question 2 - Solution

A constant-gravity model needs an explicit error convention

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Set x=h/R≥0x=h/R\ge0. With true-value normalization, ϵt=g0−gg=g0g−1=(1+x)2−1.\epsilon_t=\frac{g_0-g}{g}=\frac{g_0}{g}-1=(1+x)^2-1. The one-percent condition yields 2x+x2≤.01,x≤1.01−1=0.00498756.2x+x^2\le.01,\qquad \boxed{x\le\sqrt{1.01}-1=0.00498756}.

(b) Result and interpretation

Surface-value normalization instead gives ϵs=g0−gg0=1−(1+x)−2≤.01.\epsilon_s=\frac{g_0-g}{g_0}=1-(1+x)^{-2}\le.01. Hence x≤1.99−1=0.00503782,ϵs=ϵt1+ϵt.\boxed{x\le\frac1{\sqrt{.99}}-1=0.00503782},\qquad \epsilon_s=\frac{\epsilon_t}{1+\epsilon_t}. The larger reference denominator permits a slightly larger altitude.

Physical check

Both conventions agree to first order, ϵ≃2h/R\epsilon\simeq2h/R, but differ at the precision requested.