Question

The acceleration due to gravity at a height of $(\sqrt{2} - 1)R$ from the surface of the earth is (where $g = 10\, m/s^2$ and $R$ is the radius of the earth)

• A. \(2.5\, m/s^2\)
• B. \(7.5\, m/s^2\)
• C. \(5\, m/s^2\)
• D. \(10\, m/s^2\)

Hint:

Acceleration due to gravity decreases with height according to the inverse square law relative to the distance from Earth's center.

Answer 1

The Correct Option is C

Acceleration due to gravity at height \(h\) is given by: \[ g_h = g \left(\frac{R}{R + h}\right)^2 \] Given: \[ h = (\sqrt{2} - 1) R \] Calculate: \[ g_h = 10 \left(\frac{R}{R + (\sqrt{2} - 1) R}\right)^2 = 10 \left(\frac{1}{1 + \sqrt{2} - 1}\right)^2 = 10 \left(\frac{1}{\sqrt{2}}\right)^2 = 10 \times \frac{1}{2} = 5\, m/s^2 \]