The gravitational force of attraction between earth and the sun is 35.4710^21 N, Calculate the mass of the sun, given the mass of the earth is 5.9810^24 kg and the mean distance between the earth and the sun is 1.49610^11m given G=6.6710^-11
The Correct Answer and Explanation is:
The gravitational force of attraction between two objects can be calculated using Newton’s law of universal gravitation:F=G⋅m1⋅m2r2F = \frac{G \cdot m_1 \cdot m_2}{r^2}F=r2G⋅m1⋅m2
Where:
- FFF is the gravitational force between the two objects (in this case, between the Earth and the Sun), which is 35.47×1021 N35.47 \times 10^{21} \, \text{N}35.47×1021N,
- GGG is the gravitational constant, 6.67×10−11 N⋅m2/kg26.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^26.67×10−11N⋅m2/kg2,
- m1m_1m1 is the mass of the Earth, 5.98×1024 kg5.98 \times 10^{24} \, \text{kg}5.98×1024kg,
- m2m_2m2 is the mass of the Sun (which we need to find),
- rrr is the distance between the centers of the two bodies, 1.496×1011 m1.496 \times 10^{11} \, \text{m}1.496×1011m.
Rearrange the formula to solve for m2m_2m2, the mass of the Sun:m2=F⋅r2G⋅m1m_2 = \frac{F \cdot r^2}{G \cdot m_1}m2=G⋅m1F⋅r2
Now, plug in the known values:m2=(35.47×1021)⋅(1.496×1011)2(6.67×10−11)⋅(5.98×1024)m_2 = \frac{(35.47 \times 10^{21}) \cdot (1.496 \times 10^{11})^2}{(6.67 \times 10^{-11}) \cdot (5.98 \times 10^{24})}m2=(6.67×10−11)⋅(5.98×1024)(35.47×1021)⋅(1.496×1011)2
First, calculate the square of the distance:r2=(1.496×1011)2=2.238×1022 m2r^2 = (1.496 \times 10^{11})^2 = 2.238 \times 10^{22} \, \text{m}^2r2=(1.496×1011)2=2.238×1022m2
Now, calculate the numerator:F⋅r2=(35.47×1021)⋅(2.238×1022)=7.950×1044 N⋅m2F \cdot r^2 = (35.47 \times 10^{21}) \cdot (2.238 \times 10^{22}) = 7.950 \times 10^{44} \, \text{N} \cdot \text{m}^2F⋅r2=(35.47×1021)⋅(2.238×1022)=7.950×1044N⋅m2
Now, calculate the denominator:G⋅m1=(6.67×10−11)⋅(5.98×1024)=3.986×1014 N⋅m2/kgG \cdot m_1 = (6.67 \times 10^{-11}) \cdot (5.98 \times 10^{24}) = 3.986 \times 10^{14} \, \text{N} \cdot \text{m}^2/\text{kg}G⋅m1=(6.67×10−11)⋅(5.98×1024)=3.986×1014N⋅m2/kg
Finally, divide the numerator by the denominator to find the mass of the Sun:m2=7.950×10443.986×1014=1.994×1030 kgm_2 = \frac{7.950 \times 10^{44}}{3.986 \times 10^{14}} = 1.994 \times 10^{30} \, \text{kg}m2=3.986×10147.950×1044=1.994×1030kg
Thus, the mass of the Sun is approximately:m2=1.994×1030 kgm_2 = 1.994 \times 10^{30} \, \text{kg}m2=1.994×1030kg
This calculation shows that the Sun’s mass is around 1.994×1030 kg1.994 \times 10^{30} \, \text{kg}1.994×1030kg, which is in close agreement with the commonly accepted value of approximately 2×1030 kg2 \times 10^{30} \, \text{kg}2×1030kg.
