Respuesta :
Answer:
the force decreases by a factor of 4
Explanation:
The expression for the law of universal gravitation is
F = [tex]G \frac{m_1m_2}{r^2}[/tex]
let's call the force Fo for the distance r
F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]
They indicate that the distance doubles
r ’= 2 r
we substitute
F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]
F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]
F = ¼ F₀
consequently the correct answer is that the force decreases by a factor of 4
If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.
Let the initial force be F
Let the initial distance apart be r
Thus, we can obtain the final force as follow:
Initial force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = 2r
Final force (F₂) =?
F = GM₁M₂ / r²
Fr² = GM₁M₂ (constant)
Thus,
F₁r₁² = F₂r₂²
Fr² = F₂(2r)²
Fr² = F₂4r²
Divide both side by 4r²
F₂ = Fr² /4r²
F₂ = F / 4
From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4
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