Question

earth's gravitation field strength is 9.80n/kg. determine the distance as a multiple of earth's radii, above earth's surface which the magnitude of the acceleration due to gravity is 3.2

Answer 1

The distance above Earth's surface at which the acceleration due to gravity is 3.2 N/kg is approximately 1.78 times Earth's radius.

Step-by-step explanation:

The question asks at what multiple of Earth's radius above the surface the acceleration due to gravity is 3.2 N/kg. The gravitational field strength at a distance r from the center of the Earth is given by the formula g = GM/r^2, where G is the gravitational constant, M is Earth's mass, and r is the distance to the point of interest from the center of Earth.

To solve for r, we set g to 3.2 N/kg and solve the equation, considering the gravitational field strength at Earth's surface (g0 = 9.80 N/kg) and Earth's radius (Re = 6.371 × 10^6 m).

Therefore, we have:

3.2 = 9.80 × (Re/r)^2

Solving for r in terms of Re yields:

r = Re × √(9.80/3.2)

Completing the calculation:

r = Re × √(9.80/3.2) = 6.371 × 10^6 m × √(3.0625) ≈ 1.78Re

So, the distance above Earth's surface as a multiple of Earth's radius at which the acceleration due to gravity is 3.2 N/kg is approximately 1.78 times the Earth's radius.