1) [tex]0.61 m/s^2[/tex]
2) 13.9 s
Explanation:
1)
The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.
It can be calculated using the equation:
[tex]g=\frac{GM}{r^2}[/tex] (1)
where
G is the gravitational constant
[tex]M=5.98\cdot 10^{24} kg[/tex] is the Earth's mass
r is the distance of the object from the Earth's center
The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is
[tex]r=4R[/tex]
where
[tex]R=6.37\cdot 10^6 m[/tex] is the Earth's radius
Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):
[tex]g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2[/tex]
2)
The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where
L is the length of the pendulum
g is the acceleration due to gravity at the location of the pendulum
Note that the period of a pendulum does not depend on its mass.
For the pendulum in this problem, we have:
L = 3 m is its length
[tex]g=0.61 m/s^2[/tex] is the acceleration due to gravity (calculated in part 1)
Therefore, the period of the pendulum is:
[tex]T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s[/tex]
