If Earth's radius were reduced by 1% while its mass remained unchanged, what would be the effect on the gravitational acceleration at its surface?

Correct — C. With M fixed, g depends on radius alone:
g = GM⁄R², so g ∝ 1⁄R²

Radius falls 1%: new R = 0.99R
New g = g ÷ 0.99² = g ÷ 0.9801
New g ≈ 1.0203 g, a rise of about 2% → option (c)

Shortcut: g ∝ R⁻², so the change in g ≈ −2 × (−1%) = +2%.

Why the others are wrong

• (a)
  Increase by approximately 1% — +1% forgets the square. g varies as 1⁄R², not 1⁄R, so a 1% cut in radius lifts g by about 2%.
• (b)
  Decrease by approximately 2% — Right size, wrong direction. A smaller radius puts the surface closer to the centre of the same mass, so the pull there grows. g rises; it does not fall.
• (d)
  Decrease by approximately 1% — Wrong on both counts. It has g falling as the radius shrinks, and changing with R rather than R². The inverse square gives a rise of about 2%.

Newton's law of gravitation gives the acceleration at a planet's surface as g = GM⁄R², where M is the planet's mass and R its radius. At Earth's surface g is about 9.8 m/s².

Because R is squared, g is sensitive to radius. For small changes, the percentage change in g ≈ −2 × the percentage change in R.

The same law makes g fall with height above the surface, since the distance from the centre grows.

Key facts

• Acceleration due to gravity at a planet's surface: g = GM⁄R².
• g at Earth's surface is about 9.8 m/s².
• With mass fixed, a small percentage change in radius changes g by about twice that percentage, in the opposite direction.
• g does not depend on the mass of the falling body.

Common traps

• Dropping the square: g depends on 1⁄R², so the radius effect doubles in percentage terms.
• Getting the direction backwards: shrinking the radius with mass unchanged increases g.

No directly related past PYQ was found.