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Sagot :
Answer:
No. When the distance is doubled, the gravitational force is 1/4 times as strong, therefore, the correct answer is she is not correct
Explanation:
The gravitational force is given as follows;
[tex]F} =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}[/tex]
Where;
m₁, and m₂ are the masses sharing the gravitational field
G = The universal gravitational constant
r = The distance between the centers of the two masses
Therefore, when the new distance, R = 2 × r, we get;
[tex]F} =G \times \dfrac{m_{1} \times m_{2}}{R^{2}} = G \times \dfrac{m_{1} \times m_{2}}{(2 \times r)^{2}} = G \times \dfrac{m_{1} \times m_{2}}{4 \times r^{2}} = \dfrac{1}{4} \times G \times \dfrac{m_{1} \times m_{2}}{r^{2}}[/tex]
Therefore, when the distance is doubled, the gravitational force is one fourth as strong, therefore, she is not correct.
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