What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth? A) 4.8 m/s² B) 1.1 m/s² C) 3.3 m/s² D) 2.5 m/s² E) 6.5 m/s²

Question

Quizplus · Accepted Answer

The magnitude of the free-fall acceleration at a distance 2R from the surface of the Earth is given by the equation:
g' = (g * R^2) / (2 * (R + h)^2)
where g is the acceleration due to gravity at the surface of the Earth (9.81 m/s^2), R is the radius of the Earth (6,371 km), and h is the distance above the surface (2R).
Plugging in the values, we get:
g' = (9.81 * (6,371 km)^2) / (2 * (6,371 km + 2(6,371 km))^2)
g' ≈ 1.1 m/s^2

What Is the Magnitude of the Free-Fall Acceleration at a Point