Astronauts on the first trip to Mars take along a pendulum that has a period on earth of $1.50s$. The period on Mars turns out to be $2.45s$. What is the free-fall acceleration on Mars?

The period $T$of a simple pendulum of length $L$is given by,

$T=2\pi \sqrt{\frac{L}{g}}$

where $g$is the acceleration of gravity.

Note that the period of a simple pendulum depends only on its length and the magnitude of the gravitational constant.

It does not depend on the mass of the object hanging at its end or the amplitude of vibration.

The period of a pendulum on Earth is:: $T_{\mathrm{E}}=1.50\mathrm{s}$.

The period of the pendulum on Mars is:$T_{\mathrm{M}}=2.45\mathrm{s}$.

In part (a), we are asked to determine the period of the pendulum on the earth.

In part (b), we are asked to determine the period of the pendulum on Venus.

(a)

The period of the pendulum on the earth is found from Equation $(*)$:

$T_{\mathrm{E}}=2\pi \sqrt{\frac{L}{g_{\mathrm{E}}}}$

where $g_{\mathrm{E}}=9.80\mathrm{m}/{\mathrm{s}}^2$on the earth.

Similarly, the duration of a pendulum on Mars is:

$T_{\mathrm{M}}=2\pi \sqrt{\frac{L}{g_{\mathrm{M}}}}$

where $g_{\mathrm{M}}$is the Martian free-fall acceleration.

Divide Equation (1) by Equation (2):

$\frac{T_{\mathrm{E}}}{T_{\mathrm{M}}}=\frac{2\pi \sqrt{\frac{L}{g_{\mathrm{E}}}}}{2\pi \sqrt{\frac{L}{g_{\mathrm{M}}}}}$

$\frac{T_{\mathrm{E}}}{T_{\mathrm{M}}}=\sqrt{\frac{g_{\mathrm{M}}}{g_{\mathrm{E}}}}$

Square both sides:

$\frac{g_{\mathrm{M}}}{g_{\mathrm{E}}}={\left(\frac{T_{\mathrm{E}}}{T_{\mathrm{M}}}\right)}^2$

Solve for $g_{\mathrm{M}}$:

$g_{\mathrm{M}}=g_{\mathrm{E}}{\left(\frac{T_{\mathrm{E}}}{T_{\mathrm{M}}}\right)}^2$

Substitute numerical values:

$g_{\mathrm{M}}=\left(9.80\mathrm{m}/{\mathrm{s}}^2\right){\left(\frac{1.50\mathrm{s}}{2.45\mathrm{s}}\right)}^2$

$=3.67\mathrm{m}/{\mathrm{s}}^2$.