Physics · Gravitation · NEET
No. The force on both masses is exactly equal in magnitude, F = G m1 m2 / r squared. This is Newton's third law - the Earth pulls you with the same force you pull the Earth. What differs is the acceleration: a = F/m, so the small mass accelerates a lot and the big mass barely moves. In numericals, compute one value of F and apply it to both bodies.
Force depends on 1/r squared, so it changes with the square of the distance. Double r (r to 2r): force becomes 1/2 squared = 1/4. Triple r: force becomes 1/9. Halve r: force becomes 4 times. Always square the ratio - this is the most tested idea in NEET numericals.
r is the distance between the CENTRES of the two masses, not the surface gap and not the radius of a body. For two small balls treat them as point masses at their centres. For a body on Earth's surface, r = radius of Earth (centre to surface). Using the wrong r is the top mistake in these problems.
That is correct and expected. Because G = 6.67 x 10^-11 is so small, everyday masses feel almost no gravitational pull toward each other. A tiny answer means your setup is right. Gravity only becomes large when one mass is planet-sized (like Earth's 6 x 10^24 kg).
No. The force between mass 1 and mass 2 stays G m1 m2 / r squared no matter what sits between them - gravity cannot be blocked or shielded. A third mass adds its OWN separate pull, and you add forces as vectors, but it never cancels the original pair's force.
Two astronauts are floating in gravitational free space after losing contact with their spaceship. The two will:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
F = G m1 m2 / r squared, where G = 6.67 x 10^-11 N m squared / kg squared, m1 and m2 are the two masses in kg, and r is the distance between their centres in metres. The force is attractive and acts along the line joining the two masses.
Step 1: write down m1, m2, r in SI units (kg, m). Step 2: substitute into F = G m1 m2 / r squared. Step 3: multiply the masses, divide by r squared, then multiply by 6.67 x 10^-11. Step 4: keep the answer in scientific notation. If asked about a change in distance, just square the ratio of distances.
Yes. Gravitational force is always attractive - there is no repulsive gravity. Both masses are pulled toward each other along the line joining their centres, with equal and opposite forces.
G has units N m squared / kg squared (or m cubed / kg s squared) and dimensions [M^-1 L^3 T^-2]. Its value 6.67 x 10^-11 makes gravitational forces between ordinary objects extremely small.
Because G is only 6.67 x 10^-11, the force between two ordinary masses is around 10^-8 N or smaller - too small to feel. Gravity only becomes strong when at least one mass is huge, like a planet or star.
