If Earth’s radius shrinks by 1% with mass constant, surface gravitational acceleration would:

The correct option is Rise by about 2%.

To understand how the surface gravitational acceleration changes when Earth's radius shrinks, let us look at the formula for gravitational acceleration at the surface of a spherical mass:

$g=\frac{GM}{R^2}$

Where:
• g is the acceleration due to gravity,
• G is the universal gravitational constant,
• M is the mass of the Earth (which remains constant),
• R is the radius of the Earth.

Since G and M are constants, g is inversely proportional to the square of the radius R:

$g\∝\frac{1}{R^2}=R^{-2}$

For small percentage changes, we can use the fractional error/change approximation method by taking the relative derivative:

$\frac{\mathrm{\Δ}g}{g}=-2\×\frac{\mathrm{\Δ}R}{R}$

Multiplying by 100 to convert to percentage changes gives:

$(\frac{\mathrm{\Δ}g}{g}\×100)=-2\×(\frac{\mathrm{\Δ}R}{R}\×100)$

Given that the radius shrinks by 1%, we have:

$\frac{\mathrm{\Δ}R}{R}\×100=-1\%$

Substituting this value into the equation:

$\frac{\mathrm{\Δ}g}{g}\×100=-2\×(-1\%)=+2\%$

The positive sign indicates an increase. Therefore, the surface gravitational acceleration will rise by about 2%.