Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.081.08 m. She sets the pendulum swinging, and her collaborators carefully count 101101 complete cycles of oscillation during 2.00×1022.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer 1

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is
`g = 167.2 \ m/s^2`

Step-by-step explanation:

From the question we are told that

The length of the simple pendulum is
`L = 1.081.08 \ m`

The number of cycles is
`N = 101`

The time take is
`t = 2.00 *10^(2 \ )s`

Generally the period of this oscillation is mathematically evaluated as

`T = (N)/(t )`

substituting values

`T = (101)/(2.0*10^2 )`

`T = 0.505 \ s`

The period of this oscillation is mathematically represented as

`T = 2 \pi \sqrt{(l)/(g) }`

making g the subject of the formula we have

`g = (L)/([(T)/(2 \pi ) ]^2 )`

`g = (4 \pi ^2 L )/(T^2 )`

Substituting values

`g = (4 * 3.142 ^2 * 1.08 )/(505.505^2 )`

`g = (4 * 3.142 ^2 * 1.08 )/(0.505^2 )`

`g = 167.2 \ m/s^2`