Answer:
54.6 m/s^2
Step-by-step explanation:
length of the pendulum, l = 1 m
time period of the pendulum, T = 0.85 s
According to the formula of time period of pendulum
![T = 2\pi \sqrt{(l)/(g)}](https://img.qammunity.org/2020/formulas/physics/middle-school/dot3upfl83f4ame951ptccwl4of818e362.png)
![g = (4\pi ^(2)l)/(T^(2))](https://img.qammunity.org/2020/formulas/physics/college/yz8r18nckmkb90ndvlx0wczm3sxjx92rr8.png)
![g = (4* 3.14* 3.14* 1)/(0.85* 0.85)](https://img.qammunity.org/2020/formulas/physics/college/dxpjidygenn5v0jlfuxyalxuq0lzhk4j6t.png)
g = 54.6 m/s^2
Thus, the acceleration due to gravity at this planet is 54.6 m/s^2.