Assume that a tunnel is dug across the earth (radius = R) passing through its center. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of √(gR) (b) It is released from a height R above the tunnel (c) it is thrown vertically upward along the length of the tunnel with a speed of √(gR).
Answers
Answer ⇒ Answer of all three part is t = (π/2)√(R/g)
Explanation ⇒ All the three parts are solved below.
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(a). Speed of the particle with which it will projected inside of the earth = √(gr)
We know that the motion of the particle will be an S.H.M. since, hen tunnel is dug, and body is projected into it, it will do S.H.M. [There is an proof for that, but I don't think to give the proof here, since question is big and that fact should be learn by reader as an standard result.]
We know,
v = ⍵√(A²-x²), where A is amplitude, and x is displacement.
∴ v² = ⍵²(A² - x²)
∴ gR = g/R(A² - R²)
∴ R² = A² - R²
∴ A² = 2R²
∴ A = √2R
Now, Using the formula,
x = ASin(ωt + Φ)
For finding the value of x, first we need to find the value of Φ.
For Φ, Set t = 0,
x = ASin(Φ)
SinΦ = R/A
Φ = π/4, 3π/4 etc.
Now, which value to choose.
Always choose higher value. Do not exceeds 180 degree.
Now,
x = R√2.sin{√(g/R)t+3π/4}
Since we have taken center of the earth as origin, the displacement at other end of the tunnel x = -R
So, -R = R√2sin{√(g/R)t+3π/4}
sin{√(g/R)t+3π/4} = -1/√2
Taking Sin⁻¹ both sides,
√(g/R)t+3π/4} = Sin(-1/√2)
√(g/R)t+3π/4 = 5π/4
√(g/R)t=2π/4 = π/2
∴ t = (π/2)√(R/g)
This is the Required time.
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(b). When the particle is released from a height R above the tunnel, its velocity at the starting of the tunnel can be calculated as,
1/2 × mv² = GMm/(R+R)
∴ v² =2GM/2R
∴ v² = GM/R
Since, GM = gR²
∴ v² = gR²/R
∴ v² =gR
∴ v =√(gR)
Since, it is same as the previous value. Hence, the time period will be same as that of previous one.
Thus, t = (π/2)√(R/g)
____________________________
(c). When the particle is thrown vertically, then also at the entrance, potential energy will be same as the kinetic energy and hence the speed is same as that of √gR.
Thus, still the answer will remains unchanged.
Hence, t = (π/2)√(R/g)
____________________________
Hope it helps.
Answer:
Answer ⇒ Answer of all three part is t = (π/2)√(R/g)
Explanation ⇒ All the three parts are solved below.
____________________________
(a). Speed of the particle with which it will projected inside of the earth = √(gr)
We know that the motion of the particle will be an S.H.M. since, hen tunnel is dug, and body is projected into it, it will do S.H.M. [There is an proof for that, but I don't think to give the proof here, since question is big and that fact should be learn by reader as an standard result.]
We know,
v = ⍵√(A²-x²), where A is amplitude, and x is displacement.
∴ v² = ⍵²(A² - x²)
∴ gR = g/R(A² - R²)
∴ R² = A² - R²
∴ A² = 2R²
∴ A = √2R
Now, Using the formula,
x = ASin(ωt + Φ)
For finding the value of x, first we need to find the value of Φ.
For Φ, Set t = 0,
x = ASin(Φ)
SinΦ = R/A
Φ = π/4, 3π/4 etc.
Now, which value to choose.
Always choose higher value. Do not exceeds 180 degree.
Now,
x = R√2.sin{√(g/R)t+3π/4}
Since we have taken center of the earth as origin, the displacement at other end of the tunnel x = -R
So, -R = R√2sin{√(g/R)t+3π/4}
sin{√(g/R)t+3π/4} = -1/√2
Taking Sin⁻¹ both sides,
√(g/R)t+3π/4} = Sin(-1/√2)
√(g/R)t+3π/4 = 5π/4
√(g/R)t=2π/4 = π/2
∴ t = (π/2)√(R/g)
This is the Required time.
________________________________
(b). When the particle is released from a height R above the tunnel, its velocity at the starting of the tunnel can be calculated as,
1/2 × mv² = GMm/(R+R)
∴ v² =2GM/2R
∴ v² = GM/R
Since, GM = gR²
∴ v² = gR²/R
∴ v² =gR
∴ v =√(gR)
Since, it is same as the previous value. Hence, the time period will be same as that of previous one.
Thus, t = (π/2)√(R/g)
____________________________
(c). When the particle is thrown vertically, then also at the entrance, potential energy will be same as the kinetic energy and hence the speed is same as that of √gR.
Thus, still the answer will remains unchanged.
Hence, t = (π/2)√(R/g)
____________________________
Hope it helps.