“Find the gravitational force between the Sun and the Earth. Given Mass of the Sun = 1.99×10^30 kg Mass of the Earth = 5.98×10^24 kg The average distance between the Earth and the Sun = 1.5×10^11 m.”
The Correct Answer and Explanation is:
To find the gravitational force between the Sun and the Earth, we 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,
- GGG is the gravitational constant, 6.67×10−11 Nm2/kg26.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^26.67×10−11Nm2/kg2,
- M1M_1M1 and M2M_2M2 are the masses of the two objects (in this case, the Sun and the Earth),
- rrr is the distance between the centers of the two objects (average distance between the Earth and the Sun).
Step-by-Step Solution:
We are given:
- Mass of the Sun M1=1.99×1030 kgM_1 = 1.99 \times 10^{30} \, \text{kg}M1=1.99×1030kg,
- Mass of the Earth M2=5.98×1024 kgM_2 = 5.98 \times 10^{24} \, \text{kg}M2=5.98×1024kg,
- Distance between the Earth and the Sun r=1.5×1011 mr = 1.5 \times 10^{11} \, \text{m}r=1.5×1011m,
- Gravitational constant G=6.67×10−11 Nm2/kg2G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2G=6.67×10−11Nm2/kg2.
Substitute these values into the equation:F=(6.67×10−11)⋅(1.99×1030)⋅(5.98×1024)(1.5×1011)2F = \frac{(6.67 \times 10^{-11}) \cdot (1.99 \times 10^{30}) \cdot (5.98 \times 10^{24})}{(1.5 \times 10^{11})^2}F=(1.5×1011)2(6.67×10−11)⋅(1.99×1030)⋅(5.98×1024)
First, calculate the denominator:r2=(1.5×1011)2=2.25×1022 m2r^2 = (1.5 \times 10^{11})^2 = 2.25 \times 10^{22} \, \text{m}^2r2=(1.5×1011)2=2.25×1022m2
Now, calculate the numerator:(6.67×10−11)⋅(1.99×1030)=1.32633×1020(6.67 \times 10^{-11}) \cdot (1.99 \times 10^{30}) = 1.32633 \times 10^{20}(6.67×10−11)⋅(1.99×1030)=1.32633×10201.32633×1020⋅(5.98×1024)=7.93×10441.32633 \times 10^{20} \cdot (5.98 \times 10^{24}) = 7.93 \times 10^{44}1.32633×1020⋅(5.98×1024)=7.93×1044
Now, divide the numerator by the denominator:F=7.93×10442.25×1022=3.52×1022 NF = \frac{7.93 \times 10^{44}}{2.25 \times 10^{22}} = 3.52 \times 10^{22} \, \text{N}F=2.25×10227.93×1044=3.52×1022N
Final Answer:
The gravitational force between the Earth and the Sun is approximately:F=3.52×1022 NF = 3.52 \times 10^{22} \, \text{N}F=3.52×1022N
Explanation:
The gravitational force between two objects depends directly on their masses and inversely on the square of the distance between them. In this case, we used the known values for the masses of the Sun and Earth and the distance between them. By plugging these into the formula, we found that the gravitational force between the two objects is extremely large, demonstrating the immense pull they have on each other. This force keeps the Earth in orbit around the Sun, ensuring the stability of our solar system.
