Answer :
To find the gravitational force between the ballerina and the Earth, we will use Newton's law of universal gravitation. Newton's law of universal gravitation states that every mass exerts an attractive force on every other mass. This gravitational force [tex]\( F \)[/tex] can be calculated using the formula:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.67430 \times 10^{-11} \, m^3 \, kg^{-1} \, s^{-2})\)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the ballerina [tex]\((40 \, kg)\)[/tex],
- [tex]\( m_2 \)[/tex] is the mass of the Earth [tex]\((5.98 \times 10^{24} \, kg)\)[/tex],
- [tex]\( r \)[/tex] is the distance between the center of the ballerina and the center of the Earth [tex]\((6.38 \times 10^6 \, m)\)[/tex].
We substitute these values into the formula:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{40 \times 5.98 \times 10^{24}}{(6.38 \times 10^6)^2} \][/tex]
Let's simplify the terms inside the parentheses first:
- [tex]\( m_1 \times m_2 = 40 \times 5.98 \times 10^{24} = 239.2 \times 10^{24} \)[/tex]
- [tex]\( r^2 = (6.38 \times 10^6)^2 = 4.07044 \times 10^{13} \)[/tex]
With these, our formula now looks like:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{239.2 \times 10^{24}}{4.07044 \times 10^{13}} \][/tex]
Next, we perform the division in the fraction:
[tex]\[ \frac{239.2 \times 10^{24}}{4.07044 \times 10^{13}} \approx 5.8767 \times 10^{10} \][/tex]
Now multiply this result by [tex]\( G \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \times 5.8767 \times 10^{10} \][/tex]
The result of this multiplication gives us the gravitational force:
[tex]\[ F \approx 392.2162124979117 \, N \][/tex]
Therefore, the gravitational force between the ballerina (with a mass of [tex]\( 40 \, kg \)[/tex]) and the Earth (with a mass of [tex]\( 5.98 \times 10^{24} \, kg \)[/tex]), given the distance between them [tex]\( (6.38 \times 10^6 \, m) \)[/tex], is approximately [tex]\( 392.2162124979117 \, N \)[/tex].
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.67430 \times 10^{-11} \, m^3 \, kg^{-1} \, s^{-2})\)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the ballerina [tex]\((40 \, kg)\)[/tex],
- [tex]\( m_2 \)[/tex] is the mass of the Earth [tex]\((5.98 \times 10^{24} \, kg)\)[/tex],
- [tex]\( r \)[/tex] is the distance between the center of the ballerina and the center of the Earth [tex]\((6.38 \times 10^6 \, m)\)[/tex].
We substitute these values into the formula:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{40 \times 5.98 \times 10^{24}}{(6.38 \times 10^6)^2} \][/tex]
Let's simplify the terms inside the parentheses first:
- [tex]\( m_1 \times m_2 = 40 \times 5.98 \times 10^{24} = 239.2 \times 10^{24} \)[/tex]
- [tex]\( r^2 = (6.38 \times 10^6)^2 = 4.07044 \times 10^{13} \)[/tex]
With these, our formula now looks like:
[tex]\[ F = 6.67430 \times 10^{-11} \frac{239.2 \times 10^{24}}{4.07044 \times 10^{13}} \][/tex]
Next, we perform the division in the fraction:
[tex]\[ \frac{239.2 \times 10^{24}}{4.07044 \times 10^{13}} \approx 5.8767 \times 10^{10} \][/tex]
Now multiply this result by [tex]\( G \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \times 5.8767 \times 10^{10} \][/tex]
The result of this multiplication gives us the gravitational force:
[tex]\[ F \approx 392.2162124979117 \, N \][/tex]
Therefore, the gravitational force between the ballerina (with a mass of [tex]\( 40 \, kg \)[/tex]) and the Earth (with a mass of [tex]\( 5.98 \times 10^{24} \, kg \)[/tex]), given the distance between them [tex]\( (6.38 \times 10^6 \, m) \)[/tex], is approximately [tex]\( 392.2162124979117 \, N \)[/tex].
