Question 127: Gravitation

Question

The length of a seconds pendulum at a height h = 2R from earth surface will be:
(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $$\pi$$
2
ms
$$-$$2
)

Accepted Answer

$$g = {{GM} \over {{{(R + h)}^2}}} = {{GM} \over {9{R^2}}} = {{{g_0}} \over 9}$$
$$ \Rightarrow T = 2\pi \sqrt {{l \over g}}  = 2\pi \sqrt {{l \over {{{{g_0}} \over 9}}}} $$
$$ \Rightarrow 2 = 2\pi \sqrt {{{9l} \over {{g_0}}}} $$
$$ \Rightarrow l = {{{g_0}} \over {9{\pi ^2}}} = {1 \over 9}\,m$$

Answer

$${2 \over 9}$$ m

Answer

$${4 \over 9}$$ m

Answer

$${8 \over 9}$$ m

Answer

$${1 \over 9}$$ m

Gravitation

The length of a seconds pendulum at a height h = 2R from earth surface will be: (Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $\pi 2 ms -$2 )

Choose the correct answer: