What is the acceleration due to gravity at a distance of one Earth radius above the Earth's surface

g(Re/r)^2 = (1/4)(9.81)

Hmm, let me calculate that for you, but before that, let me check my "gravity-altimeter" here... Ah, got it! At a distance of one Earth radius above the surface, the acceleration due to gravity is approximately 8.7 meters per second squared. Just remember to wear some anti-gravity shoes if you plan on visiting so you don't float away!

The acceleration due to gravity at a distance of one Earth radius above the Earth's surface can be calculated using the formula for gravitational acceleration:

a = G * M / r^2

where:
a is the acceleration due to gravity,
G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2),
M is the mass of the Earth (approximately 5.972 × 10^24 kg),
and r is the distance from the center of the Earth to the point (in this case, one Earth radius).

The radius of the Earth is approximately 6,371 kilometers or 6,371,000 meters. Therefore, the distance from the center of the Earth to a point one Earth radius above the surface would be 2 * 6,371,000 meters.

Let's calculate the acceleration due to gravity:

a = (6.67430 x 10^-11 N m^2/kg^2) * (5.972 × 10^24 kg) / (2 * 6,371,000)^2

Simplifying the expression and calculating the values, we get:

a ≈ 9.803 m/s^2

Therefore, the acceleration due to gravity at a distance of one Earth radius above the Earth's surface is approximately 9.803 m/s^2.

To solve this question, we can use the formula for gravitational acceleration:

\(g = \frac{{G \cdot M}}{r^2}\)

where
\(g\) is the acceleration due to gravity,
\(G\) is the gravitational constant (\(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)),
\(M\) is the mass of the Earth (\(5.972 \times 10^{24} \, \text{kg}\)), and
\(r\) is the distance between the center of the Earth and the point where we want to calculate the acceleration.

In this case, since we want to find the acceleration at a distance of one Earth radius above the Earth's surface, the distance \(r\) will be the radius of the Earth (\(6.371 \times 10^6 \, \text{m}\)) plus one Earth radius (\(2 \times 6.371 \times 10^6 \, \text{m}\)).

Substituting the values into the formula:

\(g = \frac{{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \cdot 5.972 \times 10^{24} \, \text{kg}}}{{(2 \times 6.371 \times 10^6 \, \text{m})^2}}\)

Evaluating this expression will yield the acceleration due to gravity at a distance of one Earth radius above the Earth's surface.