Irodov 1.220: How High Before Gravity Drops – Amit & Deepali Sinha Foundation

Problem Irodov 1.220

Find the height above the Earth’s pole at which the acceleration of free fall has decreased by one per cent from its surface value, and separately, the height at which it has decreased by one half.

Height h above the pole, along the Earth's axis (not to scale: the same construction serves both the 1% and the 50% case, only the fraction δ differs). — Height $h$ above the pole, along the Earth’s axis (not to scale: the same construction serves both the $1\%$ and the $50\%$ case, only the fraction $\delta$ differs).

Given

$\delta = 0.01$ and $\delta = 0.50$
Earth’s radius $R$ , mass $M$

Find

The height

h

at which

g

has dropped by the fraction

\delta

Solution

Step 1Write the free-fall acceleration at the surface and at height

h

At the pole the Earth’s rotation contributes no centrifugal correction, so $g$ is simply the gravitational acceleration itself,

g_0 = \frac{\gamma M}{R^2}, \qquad g(h) = \frac{\gamma M}{(R+h)^2}. \tag{1}

Step 2Express the fractional drop and solve for

h

The problem states $g(h) = (1-\delta)\,g_0$ , so dividing the two expressions in (1),

\frac{g(h)}{g_0} = \frac{R^2}{(R+h)^2} = 1-\delta \;\;\Longrightarrow\;\; R+h = \frac{R}{\sqrt{1-\delta}}. \tag{2}

h = R\left(\dfrac{1}{\sqrt{1-\delta}}-1\right)

(3)

Step 3Evaluate for the two given fractions

With $R = 6.37\times10^6$ m,

\begin{align*} \delta=0.01: &\quad h = R\Big(\frac{1}{\sqrt{0.99}}-1\Big) \approx 0.00504\,R \approx 32\text{ km}, \\ \delta=0.50: &\quad h = R\Big(\frac{1}{\sqrt{0.50}}-1\Big) = R(\sqrt2-1) \approx 0.4142\,R \approx 2.64\times10^3\text{ km}. \tag{4}\end{align*}

h \approx 32

km for a

1\%

drop;

h \approx 2.6\times10^3

km for a

50\%

drop

Check

For small $\delta$ , $(1-\delta)^{-1/2}\approx 1+\delta/2$ , so $h\approx R\delta/2$ : at $\delta=0.01$ this linearized form gives $0.005R\approx 32$ km, matching the exact evaluation above, as it should for a $1\%$ change.

Remark

At $\delta=0.50$ , $h$ is already $41\%$ of $R$ , well past where the small- $\delta$ approximation would apply; the book’s printed $2650$ km and the $2.64\times10^3$ km found here agree to within the rounding of $R$ itself, which is the scale that dominates a change this large.