Question

Suppose we could shrink the earth without changing its mass..?At what fraction of its current radius would the free-fall acceleration at the surface be three times its present value?

Answer 1

The free-fall acceleration at the surface would be three times its present value if the radius of the earth is about 0.58 times of its current radius.

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Further explanation

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

$\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }$

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

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Given:

free fall acceleration = g' = 3g

Asked:

radius of the earth = R' = ?

Solution:

$g' : g = G \frac{M}{(R')^2} : G \frac{M}{R^2}$

$g' : g = \frac{1}{(R')^2} : \frac{1}{R^2}$

$g' : g = R^2 : (R')^2$

$3g : g = R^2 : (R')^2$

$3 : 1 = R^2 : (R')^2$

$(R')^2 = \frac{1}{3}R^2$

$R' = \sqrt{ \frac{1}{3}R^2 )$

$R' = \frac{1}{3}\sqrt{3} R$

$R' \approx 0.58 R$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer 2

Answer:

at R/$\sqrt{3}$

Explanation:

The free-fall acceleration at the surface of Earth is given by

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

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

The formula can be rewritten as

$R=\sqrt{\frac{GM}{g}}$ (1)

We want to shrink the Earth at a radius R' such that the acceleration of gravity becomes 3 times the present value, so

g' = 3g

Keeping the mass constant, M, and substituting into the equation, we have

$3g=\frac{GM}{R'^2}$

$R'=\sqrt{\frac{GM}{3g}}=\frac{1}{\sqrt{3}}\sqrt{\frac{GM}{g}}=\frac{R}{\sqrt{3}}$