A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be

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A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be

tween the satellite and the planet is f0. then the satellite is brought back to the surface of the planet. find the new force of gravitational interaction f4. express your answer in terms of f0.

Physics

2 answers:

Softa [3K] · Answer 1 · 2026-02-27T10:23:06+03:00

The relationship f4 is four times greater than f0. The gravitational interaction between two masses is defined by F = G M1M2/r^2, where F represents force, G is the gravitational constant, M1 and M2 are the respective masses, and r is the separation between their centers. In the initial scenario, the satellite's altitude is r above the surface, making its true distance from the planet's center 2r. When at the surface, the distance reduces to r. Therefore, we can calculate: f0 = G M1M2/((2r)^2) results in G M1M2/(4r^2), while f4 = G M1M2/r^2. Dividing f4 by f0 yields (G M1M2/r^2)/(G M1M2/(4r^2)), simplifying to (G M1M2/r^2) * (4r^2)/(G M1M2), ultimately leading to 4. Hence, f4 is four times f0.

inna [3.1K] · Answer 2 · 2026-02-27T10:22:11+03:00

Let M denote the mass of the planet, n refer to the mass of the satellite, and r signify the radius of the planet. When the satellite is positioned at a distance r from the planet's surface, the separation between their centers is 2r. The gravitational force acting between them can be represented by the formula $f_{0} = \frac{GMm}{(2r)^{2}} = \frac{1}{4} ( \frac{GMm}{r^{2}} )$ , where G indicates the gravitational constant. If the satellite is positioned directly on the planet's surface, the distance between the two masses becomes r, and the gravitational force is represented as $f_{4} = \frac{GMm}{r^{2}} =4f_{0}$ . The answer is: $f_{4} = 4f_{0}$ .