The gravitational force between the Sun (mass = 1.99 × 1030 kg) and Mercury (mass = 3.30 × 1023 kg) is 8.99 × 1021 N. How far is Mercury from the Sun? 6.98 × 1010 km 6.98 × 107 km 4.87 × 1013 km 4.87 × 1024 km
The Correct Answer and Explanation is:
To calculate the distance between Mercury and the Sun, we can use Newton’s Law of Universal Gravitation, which is given by: F=G⋅m1⋅m2r2F = \frac{G \cdot m_1 \cdot m_2}{r^2}F=r2G⋅m1⋅m2
Where:
- FFF is the gravitational force (8.99 × 10²¹ N),
- GGG is the gravitational constant (6.67430×10−11 Nm2/kg26.67430 \times 10^{-11} \, \text{Nm}^2/\text{kg}^26.67430×10−11Nm2/kg2),
- m1m_1m1 is the mass of the Sun (1.99 × 10³⁰ kg),
- m2m_2m2 is the mass of Mercury (3.30 × 10²³ kg),
- rrr is the distance between the two objects (Mercury and the Sun).
We need to solve for rrr (the distance). Rearranging the equation: r=G⋅m1⋅m2Fr = \sqrt{\frac{G \cdot m_1 \cdot m_2}{F}}r=FG⋅m1⋅m2
Now, plugging in the known values: r=(6.67430×10−11)⋅(1.99×1030)⋅(3.30×1023)8.99×1021r = \sqrt{\frac{(6.67430 \times 10^{-11}) \cdot (1.99 \times 10^{30}) \cdot (3.30 \times 10^{23})}{8.99 \times 10^{21}}}r=8.99×1021(6.67430×10−11)⋅(1.99×1030)⋅(3.30×1023)
First, calculate the numerator: (6.67430×10−11)⋅(1.99×1030)=1.32619×1020(6.67430 \times 10^{-11}) \cdot (1.99 \times 10^{30}) = 1.32619 \times 10^{20}(6.67430×10−11)⋅(1.99×1030)=1.32619×1020 (1.32619×1020)⋅(3.30×1023)=4.37443×1043(1.32619 \times 10^{20}) \cdot (3.30 \times 10^{23}) = 4.37443 \times 10^{43}(1.32619×1020)⋅(3.30×1023)=4.37443×1043
Now, divide by the gravitational force: 4.37443×10438.99×1021=4.86×1022\frac{4.37443 \times 10^{43}}{8.99 \times 10^{21}} = 4.86 \times 10^{22}8.99×10214.37443×1043=4.86×1022
Now, take the square root: r=4.86×1022≈6.98×1010 mr = \sqrt{4.86 \times 10^{22}} \approx 6.98 \times 10^{10} \, \text{m}r=4.86×1022≈6.98×1010m
Since 1 km = 1000 m: r≈6.98×107 kmr \approx 6.98 \times 10^{7} \, \text{km}r≈6.98×107km
Thus, the correct answer is 6.98 × 10⁷ km.
This result indicates that Mercury is about 69.8 million kilometers away from the Sun. This is a typical distance in our solar system and corresponds to Mercury’s semi-major axis, which defines its average orbital distance.
