Question

Consider the acceleration due to gravity \( g' \) at an altitude \( h \) of 50 km above the Earth's surface. If \( R \) is the radius of the Earth, and the acceleration due to gravity measured at the surface is \( g \), the ratio of \( \frac{g'}{g} \) is ________. (Assume \( h \ll R \), \( R = 6370 \, {km} \), and round off to two decimal places.)

Hint:

The acceleration due to gravity decreases with altitude. For small heights compared to Earth's radius, the decrease is approximately proportional to the square of the ratio of the radii.

Answer 1

Step 1: Using the formula for gravitational acceleration at altitude.
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by: \[ g' = g \left( \frac{R}{R + h} \right)^2 \] where:
\( g \) is the acceleration due to gravity at the surface,
\( R \) is the radius of the Earth, and
\( h \) is the height above the Earth's surface.
Step 2: Calculating the ratio \( \frac{g'}{g} \).
The ratio \( \frac{g'}{g} \) is: \[ \frac{g'}{g} = \left( \frac{R}{R + h} \right)^2 \] Substituting the given values:
\( R = 6370 \, {km} \),
\( h = 50 \, {km} \),
we get: \[ \frac{g'}{g} = \left( \frac{6370}{6370 + 50} \right)^2 = \left( \frac{6370}{6420} \right)^2 \] Step 3: Simplifying the expression.
First, calculate the ratio inside the parentheses: \[ \frac{6370}{6420} \approx 0.991 \quad \Rightarrow \quad \left( 0.991 \right)^2 \approx 0.982 \] Thus, the ratio of \( g' \) to \( g \) is approximately: \[ \frac{g'}{g} \approx 0.98 \]