A pendulum has a period of 0.70 s on Earth. What is its period on a planet where the acceleration of gravity is about 0.51 that on Earth?

Since the period is proportional to sqrt(L/g), the period will get multiplied by sqrt(1/0.51) = 1.400

That will make the new period 0.980 seconds

To determine the period of a pendulum on a different planet, where the acceleration due to gravity is different, we need to use the formula for the period of a simple pendulum:

T = 2π√(L / g)

Where:
T is the period of the pendulum,
π (pi) is a mathematical constant approximately equal to 3.14,
L is the length of the pendulum,
g is the acceleration due to gravity.

In this case, we are given the period of the pendulum on Earth (T_earth = 0.70 s) and the ratio of the acceleration due to gravity on the other planet to that on Earth (0.51). Let's assume the length of the pendulum remains the same on both Earth and the other planet.

Here's how we can find the period on the other planet:

1. First, rearrange the formula to solve for g:
g = (4π²L) / T²

2. Plug in the values for the length (L) and period (T) on Earth to find the acceleration due to gravity on Earth (g_earth):

g_earth = (4π²L) / T²_earth

3. Use the ratio of the acceleration due to gravity on the other planet to that on Earth to find the acceleration due to gravity on the other planet (g_other):

g_other = g_earth * 0.51

4. Rearrange the formula again and solve for the period on the other planet (T_other):

T_other = 2π√(L / g_other)

5. Plug in the known values of length (L) and acceleration due to gravity on the other planet (g_other) into the formula and calculate T_other.

Now, let's calculate the period on the other planet:

T_earth = 0.70 s (given)
T_earth² = 0.70² = 0.49 s²

g_earth = (4π²L) / T_earth²

T_other = 2π√(L / g_other)

Let's assume the length of the pendulum (L) is 1 meter.

g_earth = (4π² * 1) / 0.49 ≈ 39.50 m/s²

g_other = g_earth * 0.51 ≈ 39.50 * 0.51 ≈ 20.15 m/s²

T_other = 2π√(1 / 20.15) ≈ 2π√(0.0496) ≈ 2π * 0.222 ≈ 1.40 s

Therefore, the period of the pendulum on the planet with the acceleration due to gravity approximately 0.51 times that on Earth is 1.40 seconds.