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As the other contributor mentioned, the standard formula for the period (T) of a simple pendulum is

T = 2*pi*sqrt(L/g)

so the period is inversely proportional to the square root of acceleration 'g'. But for practical purposes (as implied by the question) we can replace 'g' with another value, the apparent acceleration due to gravity, 'ga'. This value also takes into account the rotational speed and the distance from the center of the gravitational mass

ga = GM/r**2 - (w**2)*cos(LAT)

where:

w = angular velocity of the earth's rotation

= 2*pi/(24*3600) [rad/s]

LAT = observer's latitude (0=equator, 90deg=pole)

G = universal gravitation constant

M, r = mass, radius of planet/satellite/star we are on

Thus, the period of a simple pendulum is inversely proportional to to the sqare root of 'ga'. And this value varies with latitude, mass and distance.

So then let's answer the questions!

a) as we increase the height from sea level, the radius increases, reducing the 'ga' and this increases the period, T

b) as we go to the pole, LAT = 90deg, and cos(LAT) goes to zero. We thus INCREASE 'ga' and decrease the period

c) at the equator, LAT = 0 and cos(LAT) = 1, so we have a minimum value for 'ga', this increases the period

d) on the moon, our rotational velocity is much less (1 rev per 27.3 days) and the M is much smaller, and the r is much smaller! We are told that the 'ga' will be about 1/6 of the Earth's value, so the period will increase.

e) Here the M is colossal, so if we could withstand the heat and gravitational forces, 'ga' is much larger, so period will decrease.

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Q: How would the period of a simple pendulum be changed if the pendulum were moved from sea level to the sun?
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