What is the gravitational force between the Earth and Mars? Mass of Earth = 5.97 x 10^24 kg Mass of Mars = 6.39 x 10^23 kg Distance between Earth and Mars = 2.25 x 10^11 m Options: A) 5.02 x 10^15 N B) 1.13 x 10^27 N C) 7.52 x 10^25 N D) 2.54 x 10^38 N
The Correct Answer and Explanation is:
To calculate the gravitational force between the Earth and Mars, we use 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,
- GGG is the gravitational constant, G=6.67×10−11 N⋅m2/kg2G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2G=6.67×10−11N⋅m2/kg2,
- m1m_1m1 and m2m_2m2 are the masses of Earth and Mars, respectively,
- rrr is the distance between the centers of Earth and Mars.
Given:
- Mass of Earth, m1=5.97×1024 kgm_1 = 5.97 \times 10^{24} \, \text{kg}m1=5.97×1024kg,
- Mass of Mars, m2=6.39×1023 kgm_2 = 6.39 \times 10^{23} \, \text{kg}m2=6.39×1023kg,
- Distance between Earth and Mars, r=2.25×1011 mr = 2.25 \times 10^{11} \, \text{m}r=2.25×1011m.
Now, substituting the values into the formula:F=(6.67×10−11)⋅(5.97×1024)⋅(6.39×1023)(2.25×1011)2F = \frac{{(6.67 \times 10^{-11}) \cdot (5.97 \times 10^{24}) \cdot (6.39 \times 10^{23})}}{{(2.25 \times 10^{11})^2}}F=(2.25×1011)2(6.67×10−11)⋅(5.97×1024)⋅(6.39×1023)
First, calculate the denominator:r2=(2.25×1011)2=5.0625×1022 m2r^2 = (2.25 \times 10^{11})^2 = 5.0625 \times 10^{22} \, \text{m}^2r2=(2.25×1011)2=5.0625×1022m2
Now, calculate the numerator:(6.67×10−11)⋅(5.97×1024)⋅(6.39×1023)=2.54×1037(6.67 \times 10^{-11}) \cdot (5.97 \times 10^{24}) \cdot (6.39 \times 10^{23}) = 2.54 \times 10^{37}(6.67×10−11)⋅(5.97×1024)⋅(6.39×1023)=2.54×1037
Finally, divide the numerator by the denominator:F=2.54×10375.0625×1022=5.02×1015 NF = \frac{{2.54 \times 10^{37}}}{{5.0625 \times 10^{22}}} = 5.02 \times 10^{15} \, \text{N}F=5.0625×10222.54×1037=5.02×1015N
Therefore, the gravitational force between the Earth and Mars is 5.02 x 10^15 N, which corresponds to option A.
