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Sergeu [11.5K]
3 years ago
8

If a fixed length simple pendulum is found to have three times the period on an unknown planet’s surface (compared to Earth), wh

at is the acceleration due to gravity on that planet? Show your work.
Physics
1 answer:
balu736 [363]3 years ago
6 0

Answer:

g/9

Explanation:

length of the pendulum = L

time period on the earth = T

Time period on the planet = 3T

Let the acceleration due to gravity on the earth is g and on the planet is g'.

Use the formula for the time period of a simple pendulum for the time period on earth

T=2\pi \sqrt{\frac{L}{g}}     .... (1)

Time period on the surface of planet is

3T=2\pi \sqrt{\frac{L}{g'}}      .... (2)

Divide equation (2) by equation (1)

\frac{3T}{T}= \sqrt{\frac{g}{g'}}

g' = g/9

Thus, the acceleration due to gravity on the planet is g /9

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