Question

A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height R above the earth's surface , where R is the radius of the earth . Then the value T2/T1 is​

Answer 1

it is the correct answer of above question

Answer 2

Answer:

The radius of the earth . Then the value T2/T1 is​ 2

Explanation:

Step 1: The Earth's radius, as measured by 183 high school classes involved in the World Year of Physics initiative "Measure the Earth with Shadows," is 6563 km, as opposed to the commonly accepted figure of 6371 km for the mean radius.

Entry-level encyclopaedia vocabulary The extremely hot and dense core of our globe is called the Earth's core. Underneath the cool, brittle crust and the largely solid mantle is the ball-shaped centre. The core has a radius of about 3,485 kilometers and is located about 2,900 kilometres (1,802 miles) below the surface of the Planet (2,165 miles).

Step 2: The equatorial diameter is marginally larger at 7930 miles, while the distance between the north and south poles is roughly 7900 miles. Simply multiplying the Earth's perimeter by pi, or 3.14159, yields its average diameter, or 7915 miles. This tells us a circumference of the Earth of about 25,000 miles.

Simple pendulum's time span$=2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$

$\Rightarrow \mathrm{g} \text { on earth's surface }=\frac{\mathrm{GM}}{\mathrm{R}^2}$

g at a height R above the surface of the planet $=\frac{\mathrm{GM}}{(2 \mathrm{R})^2}$

T 1 (epoch on the surface of the planet) $=2 \pi \sqrt{\frac{1\left(\mathrm{R}^2\right)}{\mathrm{GM}}}$

T 2 (duration at an altitude of 'h'  above the surface of the planet $=2 \pi \sqrt{\frac{1(2 \mathrm{R})^2}{\mathrm{GM}}}$

$\Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\sqrt{4}=2$

The radius of the earth . Then the value T2/T1 is​ $2$

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