A rocket moves away from the surface of Earth. As the distance r from the center of the planet increases, what happens to the force of gravity on the rocket?

Question

BlackTom AI · Accepted Answer

Newton's law of universal gravitation: F = G m1 m2 / r^2, so the gravitational force between two point masses (or a spherical mass and an external point) falls off as the inverse square of the separation r.

1) Relevant knowledge points and concepts
- Newton’s universal law: F = G M_E m / r^2, where G is the gravitational constant, M_E is Earth’s mass, m is the rocket’s mass, and r is the distance from Earth’s center.
- For a spherically symmetric mass distribution (like Earth), the external field behaves as if all mass were concentrated at the center (valid for r ≥ R_E).
- “Directly proportional to 1/r^2” means F ∝ 1/r^2.

2) Complete reasoning step by step
- Start from the law: F(r) = G M_E m / r^2.
- Treat G, M_E, and m as constants for the rocket’s flight; r is the variable.
- Therefore F(r) changes with r according to the 1/r^2 factor.

3) Intermediate calculation examples
- If r doubles (r2 = 2 r1): F2 = G M m / (2 r1)^2 = (1/4) G M m / r1^2 = (1/4) F1.
- If r triples: F3 = (1/9) F1.
These show the inverse-square decrease numerically.

4) Why the correct answer is correct
- The algebra from the universal law directly yields F ∝ 1/r^2, so as r increases the force decreases like the inverse square.

5) Why other options are incorrect
- a (F ∝ r): would mean force increases linearly with distance — contradicts the 1/r^2 law.
- b (force becomes zero after losing contact): incorrect — gravity acts at a distance and only goes to zero at infinite separation.
- d (F ∝ r^2): would mean force grows with distance, opposite of reality.
- e (force doesn’t change): false because F clearly depends on r via the denominator r^2.

6) Conclusion
The correct proportionality is F ∝ 1/r^2, which matches answer c.

A rocket moves away from the surface of Earth. As the distance r from the center of the planet increases, what happens to the force of gravity on the rocket?Single choice

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