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:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
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:
[tex]g' : g = G \frac{M}{(R')^2} : G \frac{M}{R^2}[/tex]
[tex]g' : g = \frac{1}{(R')^2} : \frac{1}{R^2}[/tex]
[tex]g' : g = R^2 : (R')^2[/tex]
[tex]3g : g = R^2 : (R')^2[/tex]
[tex]3 : 1 = R^2 : (R')^2[/tex]
[tex](R')^2 = \frac{1}{3}R^2[/tex]
[tex]R' = \sqrt{ \frac{1}{3}R^2 )[/tex]
[tex]R' = \frac{1}{3}\sqrt{3} R[/tex]
[tex]R' \approx 0.58 R[/tex]
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Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://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
