Question

The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.

Answer 1

we know, acceleration due to gravity, g = GM/R²
where M is mass of the planet , R is the radius of the planet.
so, g is directly proportional to M and inversely proportional to r².

a/c to question, mass and radius of a planet are doubled that of the earth.
e.g., M' = 2M and , R' = 2R
so, g' = G(2M)/(2R)² = GM/2R² = g/2

hence, acceleration due to gravity of planet is half of acceleration due to gravity of earth.

as time period is inversely proportional to square root of acceleration due to gravity.
so, $\frac{T_1}{T_2}=\sqrt{\frac{g_2}{g_1}}$
Where $T_1$ denotes time period of pendulum on the earth and $T_2$ denotes the time period of pendulum on the planet.

given, $T_1=T,g_1=g,g_2=g/2$

so, T/$T_2$ = √{1/2}[/tex]

$T_2$ = √2T

Answer 2

Answer:

Here is your answer

Explanation:

4.9 m/s^2