Question

If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth's surface would

Answer 1

Step 1. Given data:

Let $R$= Old radius.

$R\text{'}$= new radius

Then, $R\text{'}=R-0.01R=0.99R$…………$\left(1\right)$

Mass $M$ remains unchanged.

Step 2: Calculations:

Acceleration due to gravity: The acceleration of a body in free fall under the influence of earth's gravity is expressed as the rate of increase of velocity per unit of time and assigned the standard value of $9.81\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$. It is denoted by $g$.

We know $g=\frac{GM}{R^2}$,………………..$\left(1a\right)$ where $g$=accelration due to gravity, $G$=universal gravitational constant,$M$= mass of earth,$R$= radius of the earth.

Let $g\text{'}$= new acceleration due to gravity.

Then $g\text{'}=\frac{GM}{{\left(R\text{'}\right)}^2}$……………$\left(2\right)$

Putting the value of $R\text{'}$from $\left(1\right)$ into $\left(2\right)$….we get.

$g\text{'}=\frac{GM}{{\left(0.99R\right)}^2}=1.02\frac{GM}{R^2}$………….$\left(3\right)$

Substracting $\left(1a\right)$ by $\left(3\right)$, we get

$g\text{'}-g=\frac{GM}{R^2}(1.02-1)=0.02\frac{GM}{R^2}$

So percentage increase in $g$= $\frac{\left(g\text{'}-g\right)}{g}\times 100$

So percentage increase in $g=0.02\times 100=2\%$

Hence, the acceleration due to gravity on the earth's surface would increase by $2\%$.