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If the radius of the earth were to shrink by $1$% while its mass remained the same,

then how would the acceleration due to gravity on the earth's surface increase by $2$%?

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closed as off-topic by Thomas Fritsch, Jon Custer, John Rennie, G. Smith, Yashas May 31 at 16:25

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The initial acceleration of a mass on the Earth's surface would have the value $$ a_i = \frac{GM}{r^2}. $$ Where $M$ is the mass of the Earth and $r$ the distance from the Earth's center. If the Earth shrinks by $1 \%$ then the final acceleration will be $$ a_f = \frac{GM}{(0.99r)^2}. $$ The increase in acceleration will be equal to $ a_f/a_i = (0.99)^{-2} \simeq 1.02$. Which is an increase of approximately $2 \%$.

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I’m not entirely sure what you mean by the question, but from what it sounds like, if you shrink the radius of the earth by 1%, then objects on the surface of this new earth will be closer towards the center. Newton’s law of gravitation is an inverse-square law, in which the closer you are to the gravitating object, the greater the force will be on you. Which means that if you’re closer to the object, you’ll accelerate faster.

The equation is: $$F=\frac{Gm_1m_2}{r^2}$$

Where F is the gravitational force, $m_1$ and $m_2$ are the masses of the two gravitationally interacting objects, $r$ is the distance between their centre of masses, and $G$ is the gravitational constant, equal to about $6.67*10^{-11}$.

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