The period of a pendulum is 0.24 s on Earth. The period of the same pendulum is found to be 0.4

step 1 This question covers the concept of the time period for pendulum and that is given by 2 pi into

The period of a pendulum is 0.24 s on Earth. The period of...

The period of a pendulum is $0.24 \mathrm{~s}$ on Earth. The period of the same pendulum is found to be 0.48 s on planet $X,$ whose mass is equal to that of Earth. (a) Calculate the gravitational acceleration at the surface of planet $X$. (b) Find the radius of planet $\mathrm{X}$ in terms of that of Earth.

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Key Concepts

Simple Pendulum Period

This concept involves the formula T = 2??(L/g), which relates the period of a simple pendulum to its length and the gravitational acceleration. This relationship is crucial for understanding how changes in gravity affect the oscillation period of a pendulum, and it serves as a basic tool for measuring gravitational acceleration on different bodies.

Gravitational Acceleration

Gravitational acceleration at the surface of a celestial body is given by g = GM/R², where G is the gravitational constant, M is the mass, and R is the radius of the planet. This relationship shows that for a fixed mass, an increase in the planetary radius results in a decrease in surface gravity, which is essential for comparing gravitational fields of different planets.

Scaling Laws in Gravitational Fields

Scaling laws help relate changes in physical quantities; here, the squared period of a pendulum is inversely proportional to the gravitational acceleration. By comparing the period of a pendulum on two different planets, one can deduce how the gravitational acceleration scales, and with constant mass, infer the relative radii of the planets using the relation between mass, radius, and surface gravity.

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