Answer to Question #282194 in Mechanics | Relativity for umaima

Question

Astronauts on the first trip to Mars take along a pendulum that

has a period on earth of 1.50 s. The period on Mars turns out to be

2.45 s. What is the free-fall acceleration on Mars?

Answer 1

Solution:

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

g at earth is:

$$g=9.81\frac{m}{s^2}$$

$$\frac{T_{mars}}{T_{earth}}=\sqrt{\frac{g_{earth}}{g_{mars}}}$$

$$\implies \frac{2.45}{1.50}=\sqrt{\frac{9.81}{g_{mars}}}\implies g_{mars}=3.677\frac{m}{s^2}$$

Answer:

$$g_{mars}=3.677\frac{m}{s^2}$$