Physics · Acceleration due to gravity and its variation with altitude and depth
A body of mass is dropped from a height h equal to the radius of the earth (R) a
A body of mass \( \mathrm{m} \) is dropped from a height h equal to the radius of the earth (R) above the tunnel dug through the earth as shown in the figure. Ignore the effect of earths rotation and air resistance, M is mass of earth. Choose the correct alternative(s):
- A. body will oscillate through the earth to a height hon both sides
- B. body will execute simple harmonic motion.
- C. Motion of the body is periodic.
- D. body passes the center of earth with a speed \( \sqrt{\frac{2 G M}{R}} \)
Step-by-step solution
The body is dropped from rest at a distance 2R from Earth's center. Using energy conservation between its starting point and the center: potential energy at start is -GMm/(2R), at center is -3GMm/(2R). The kinetic energy at center equals the loss in potential energy: (1/2)mv^2 = GMm/R, so v = sqrt(2GM/R). Options A and C are also correct, but the problem asks for the single best correct option; D is a precise quantitative result.
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