The planet's radius is about 22070 km
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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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Complete Question:
You have been visiting a distant planet. your measurements have determined that the planet's mass is three times that of earth but the free-fall acceleration at the surface is only one-fourth as large. What is the planet’s radius?
Given:
free fall acceleration of the earth = g
free fall acceleration of the planet = g' = ¹/₄ g
mass of the earth = M
mass of the planet = M' = 3M
radius of the earth = 6 371 km
Asked:
radius of the planet = R' = ?
Solution:
[tex]g' : g = G \frac{M'}{(R')^2} : G \frac{M}{R^2}[/tex]
[tex]g' : g = \frac{M'}{(R')^2} : \frac{M}{R^2}[/tex]
[tex]g' : g = \frac{3M}{(R')^2} : \frac{M}{R^2}[/tex]
[tex]\frac{1}{4}g : g = 3R^2 : (R')^2[/tex]
[tex]\frac{1}{4} : 1 = 3R^2 : (R')^2[/tex]
[tex]1 : 4 = 3R^2 : (R')^2[/tex]
[tex](R')^2 = 12R^2[/tex]
[tex]R' = \sqrt{ 12R^2}[/tex]
[tex]R' = 2\sqrt{3}R[/tex]
[tex]R' = 2\sqrt{3}(6371)[/tex]
[tex]R' \approx 22070 \texttt{km}[/tex]
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Grade: High School
Subject: Physics
Chapter: Gravitational Fields
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
