Question 89: Gravitation

Question

If $\mathrm{R}$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $\mathrm{h}=2 R$ from the surface of earth will be, :

Accepted Answer

To find the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$, is related to the gravitational acceleration, $g$, and the period, $T$, by the formula: $$ T = 2\pi\sqrt{\frac{L}{g}} $$ Since we are talking about a second's pendulum, the period, $T$, is 2 seconds (since it takes one second for the pendulum to swing in one direction and another second to swing back), thus $T = 2 	ext{ seconds}$.
Now let's find the gravitational acceleration at height $h = 2R$ where $R$ is the radius of the earth. The general formula for gravitational acceleration at a height $h$ above the surface is: $$ g_h = \frac{g}{{\left(1 + \frac{h}{R}ight)}^2} $$ Plugging $h = 2R$ into the formula, we get:
$$ g_h = \frac{g}{{(1 + \frac{2R}{R})}^2} $$
$$ g_h = \frac{g}{{(1 + 2)}^2} $$
$$ g_h = \frac{g}{3^2} = \frac{g}{9} $$
So the gravitational acceleration at height $h$ is one-ninth of the gravitational acceleration at the surface of the Earth. Given that $g = \pi^2 	ext{ m/s}^2$, we get: $$ g_h = \frac{\pi^2}{9} 	ext{ m/s}^2 $$
Now knowing the gravitational acceleration at height $h$ and with the period $T$ of 2 seconds, we can rearrange the formula for the second's pendulum to solve for the length $L_h$:
$$ 2 = 2\pi\sqrt{\frac{L_h}{g_h}} $$
$$ 1 = \pi\sqrt{\frac{L_h}{g_h}} $$
$$ \frac{1}{\pi} = \sqrt{\frac{L_h}{g_h}} $$
Squaring both sides, we get:
$$ \frac{1}{\pi^2} = \frac{L_h}{g_h} $$
Multiplying both sides by $g_h$ gives us the length $L_h$:
$$ L_h = \frac{g_h}{\pi^2} $$
Substituting $g_h$ into the equation yields:
$$ L_h = \frac{\pi^2}{9\pi^2} $$
$$ L_h = \frac{1}{9} 	ext{ m} $$
Therefore, the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth is $\frac{1}{9}$ meters. The correct answer is Option A.

Answer

$\frac{1}{9} \mathrm{~m}$

Answer

$\frac{8}{9} \mathrm{~m}$

Answer

$\frac{2}{9} \mathrm{~m}$

Answer

$\frac{4}{9} \mathrm{~m}$

Gravitation

If \mathrm{R} is the radius of the earth and the acceleration due to gravity on the surface of earth is g=\pi^2 \mathrm{~m} / \mathrm{s}^2, then the length of the second's pendulum at a height \mathrm{h}=2 R from the surface of earth will be, :

Choose the correct answer: