If an object at a distance 6400 km from the center of Earth weighs 10 N, what will be its weight if it is shifted to 12,800 km from the center of Earth? [Ans: 25 N]
The Correct Answer and Explanation is:
Correct Answer: 2.5 N
Explanation:
The weight of an object is the gravitational force exerted on it by the Earth. This force is governed by Newton’s Law of Universal Gravitation, which is mathematically expressed as:F=GMmr2F = \frac{G M m}{r^2}F=r2GMm
Where:
- FFF is the gravitational force (or weight),
- GGG is the universal gravitational constant,
- MMM is the mass of the Earth,
- mmm is the mass of the object,
- rrr is the distance from the center of the Earth.
From the equation, gravitational force is inversely proportional to the square of the distance from the center of the Earth:F∝1r2F \propto \frac{1}{r^2}F∝r21
Let the initial distance be r1=6400r_1 = 6400r1=6400 km and the final distance be r2=12,800r_2 = 12,800r2=12,800 km.
Let the initial weight be F1=10F_1 = 10F1=10 N and the final weight be F2F_2F2.
Using the inverse-square law:F2F1=(r1r2)2\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^2F1F2=(r2r1)2
Substitute values:F210=(640012800)2=(12)2=14\frac{F_2}{10} = \left(\frac{6400}{12800}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}10F2=(128006400)2=(21)2=41
Solving for F2F_2F2:F2=10×14=2.5 NF_2 = 10 \times \frac{1}{4} = 2.5 \text{ N}F2=10×41=2.5 N
Thus, the weight of the object becomes 2.5 N when moved to a distance of 12,800 km from Earth’s center.
The change in weight arises because gravitational force decreases rapidly with distance. Doubling the distance from the Earth’s center causes the weight to reduce to one-fourth of its original value, demonstrating the significant effect of radial distance in gravitational interactions.
