A rocket is fired from the earth's surface to a distant planet. By Newton's law of universal gravitation, the force F will

Analyze the question: A rocket leaves the Earth's surface and travels toward a distant planet. We're asked how the gravitational force F behaves according to Newton's law of universal gravitation.

Concept: Newton's law of universal gravitation states:

F = G·m₁·m₂ / r²

• G is the universal gravitational constant. It never changes.
• m₁ and m₂ are the masses (the rocket and the body attracting it).
• r is the distance between the centres of the two bodies.

Force is inversely proportional to the square of the distance: F ∝ 1/r².

Solution:

1. Option B is wrong because G is a constant and does not vary.
2. Option C is wrong because r changes as the rocket moves, so F changes.
3. Option D is wrong because F decreases as r increases, not increases.
4. Option A says F increases as r reduces. This matches F ∝ 1/r²: when the distance between two bodies shrinks, the force between them grows.

Note on the wording: The question is a bit loosely phrased. As the rocket leaves Earth, its distance from Earth increases, so Earth's pull on it weakens. But as it approaches the distant planet, its distance from that planet reduces, so that planet's pull on it increases. The only statement consistent with the law is that F increases as r reduces, which is option A.

Answer: A

Tip: For any inverse-square law (gravity, Coulomb's law, light intensity), remember "closer means stronger, farther means weaker." If the distance doubles, the force becomes ¼; if the distance halves, the force becomes 4 times as large.