Question

10. A ball is attached to a string of length 4 m to make a pendulum. The pendulum is placed at a location that is away from the Earth’s surface by twice the radius of the Earth. What is the acceleration due to gravity at that height and what is the period of the oscillations?

Please show all of your work.

Answer 1

Let's see

• Radius of earth=6371m
• Twice of it=6371(2)=12742m

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

`\\ \rm\rightarrowtail T=2\pi \sqrt{(4)/(g)}`

`\\ \rm\rightarrowtail T=2\pi(2)/(√(g))`

`\\ \rm\rightarrowtail T=(4\pi)/(√(g))`

So

`\\ \rm\rightarrowtail g=(16\pi ^2)/(T^2)`

Answer 2

g' = 1.09 m/s²

T = 12 s

Step-by-step explanation:

Gravity due to acceleration

• Distance from center of Earth = r + 2r = 3r
• gravity due to acceleration on surface, g = GM/r²
• Substitute 3r in the denominator
• g' = GM/(3r)² = GM/9r²
• ⇒ g' = g/9 = 9.8/9 = 1.09 m/s²

Time period

• T = 2π√L/g'
• T = 2π√4/1.09
• T ≅ 12 s