Answer:
r = 0.692m
Explanation:
According to law of gravitation, the force of attraction F existing between two masses m1 and m2 is directly proportional to the product of the masses and inversely proportional to the square of the distance (r) between them.
Mathematically, F= Gm1m2/r² where G is the gravitational constant.
Given F = 8.92 x 10^-11 N
m1 = m2 = 0.8kg
G = 6.67×10^-11 m³/kgs²
r = ?
Substituting this values in the given formula, we have;
8.92×10^-11 = 6.67×10^-11(0.8)(0.8)/r²
Cross multiplying;
8.92×10^-11r² = 4.27×10^-11
r² = 4.27×10^-11/8.92×10^-11
r² = 0.479×10^0
r² = 0.479
r = √0.479
r = 0.692m
This shows that the masses are 0.692m apart.
The distance apart between the two balls is 0.69 m
From the question given above, the following data were obtained:
Mass of 1st ball (M₁) = 0.8 Kg
Mass of 2nd ball (M₂) = 0.8 Kg
Force (F) = 8.92×10¯¹¹ N
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
The distance apart between the two balls can be obtained as follow:
[tex]F = \frac{GM_1M_2}{ {r}^{2}} \\ \\ 8.92 \times {10}^{ - 11} = \frac{ 6.67 \times {10}^{ - 11} \times 0.8 \times 0.8 }{{r}^{2}} \\ \\ cross \: multiply \\ \\ 8.92 \times {10}^{ - 11} \times {r}^{2} = 6.67 \times {10}^{ - 11} \times 0.8 \times 0.8 \\ \\ divide \: both \: side \: by \: 8.92 \times {10}^{ - 11} \\ \\ {r}^{2} = \frac{6.67 \times {10}^{ - 11} \times 0.8 \times 0.8 }{8.92 \times {10}^{ - 11} } \\ \\ take \: the \: square \: root \: of \: both \: side \\ \\ r = \sqrt{ \frac{6.67 \times {10}^{ - 11} \times 0.8 \times 0.8 }{8.92 \times {10}^{ - 11}}} \\ \\ r = 0.69 \: m[/tex]
Thus, the distance apart is 0.69 m.
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