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SHM · Lecture 24 of 26

Drop a ball into a tunnel through the Earth: it does SHM

By Lalit, IIT Madras

A tunnel through the Earth as an SHM for JEE Main and Advanced: g inside the Earth gives F = −kx, and any chord tunnel keeps the same T.

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A chalkboard titled 'A tunnel through the Earth' shows a circle for the Earth with a straight tunnel along a diameter and a ball marked at a distance R/2 from the centre, beside the boxed formula T = 2π√(R/g).

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What this lecture covers

Inside a uniform solid Earth, g at radius r is g r/R, so a ball in a tunnel through the centre feels F = −kx. With g = GM/R², the period works out for any uniform planet: about 84 minutes for the Earth, so crossing takes half that. The speed at the centre comes from energy, using the potential −3GM/2R, and again from SHM. A chord tunnel gives the same T, and the wall's push on the ball stays constant.

In this lecture

  1. Play from Gravity inside a uniform Earth
  2. Play from The centre is equilibrium; F = −kx
  3. Play from T for any uniform planet, and the Earth
  4. Play from The release point does not matter
  5. Play from The time to cross the tunnel
  6. Play from Speed at the centre: energy vs SHM
  7. Play from The speed at any r, and halfway
  8. Play from A tunnel along a chord: restoring force
  9. Play from Any chord gives the same T
  10. Play from The wall's push on the chord, recap

Check yourself

  1. Asked at play from

    Question 1: For a uniform solid Earth of radius R with acceleration due to gravity g at its surface, what is the acceleration due to gravity at a distance r from the centre, inside the Earth?

    1. g r/R
    2. g R/r
    3. g R²/r²
    Show the answerHide the answer

    Answer: g r/R

  2. Asked at play from

    Question 2: A ball is in a straight tunnel along a diameter of a uniform Earth of radius R, at displacement x from the centre. Over what range of x is the gravitational force on it exactly F = −kx with k = mg/R?

    1. Every x inside the Earth
    2. Only for very small x
    3. For no x
    Show the answerHide the answer

    Answer: Every x inside the Earth

  3. Asked at play from

    Question 3: If G is the gravitational constant, M is the mass of the Earth and R is its radius, what is the acceleration due to gravity g at the Earth's surface?

    1. G M / R²
    2. G M / R
    3. G M R²
    Show the answerHide the answer

    Answer: G M / R²

  4. Asked at play from

    Question 4: A ball released from the surface oscillates in a tunnel through the centre of a uniform Earth of radius R with time period T. What happens to T if it is released from a point R/2 from the centre instead?

    1. Half
    2. Same
    3. Double
    Show the answerHide the answer

    Answer: Same

  5. Asked at play from

    Question 5: A ball released at one end of a straight tunnel through the centre of a uniform Earth oscillates with a time period of about 84 minutes. How long does it take to travel from the surface at one end to the surface at the other end?

    1. 84 minutes
    2. 42 minutes
    3. 21 minutes
    Show the answerHide the answer

    Answer: 42 minutes

  6. Asked at play from

    Question 6: Inside a uniform Earth of mass M and radius R, the gravitational potential at a distance r from the centre is −GM(3R² − r²)/2R³, which equals −GM/R at the surface. What is the potential at the centre?

    1. −3GM/2R
    2. −GM/R
    3. 0
    Show the answerHide the answer

    Answer: −3GM/2R

  7. Asked at play from

    Question 7: A ball released from rest at the surface of a uniform Earth of radius R falls through a tunnel along a diameter and has a speed of 8 km/s at the centre. What is its speed when it is at a distance R/2 from the centre?

    1. 4 km/s
    2. 6.9 km/s
    3. 8 km/s
    Show the answerHide the answer

    Answer: 6.9 km/s

  8. Asked at play from

    Question 8: A ball released at the surface end of a tunnel along a diameter of a uniform Earth oscillates with a period of about 84 minutes. Another ball is released from the surface end of a smooth straight tunnel along a chord at perpendicular distance D from the centre. How does its period compare?

    1. Shorter than 84 minutes
    2. Same, 84 minutes
    3. Longer than 84 minutes
    Show the answerHide the answer

    Answer: Same, 84 minutes

  9. Asked at play from

    Question 9: A ball slides in a smooth straight tunnel dug along a chord at perpendicular distance D from the centre of a uniform Earth of radius R. If x is the ball's displacement along the tunnel from its midpoint, what does the graph of the normal force N from the tunnel wall against x look like?

    1. A horizontal line
    2. A straight line rising with x
    3. A curve falling with x
    Show the answerHide the answer

    Answer: A horizontal line

Read the lectureIn Roman Hinglish · about 2,000 words

Play from Gravity inside a uniform Earth

Aaj gravitation aur SHM ek saath aate hain. Gravitation wale chapter mein humne Earth ke andar g padha tha. Earth ki radius R hai aur surface par g hai. Ab Earth ke andar, centre se r doori par, g kitna hota hai?

Sahi. Centre par r zero hai, toh g bhi zero ho jaata hai. Surface par r hota hai R, toh poora g waapas mil jaata hai. Yaani andar g, r ke saath badhta hai, aur r ke directly proportional rehta hai.

g × r/R, ye formula tabhi chalta hai jab Earth uniform solid ball ho. Hollow shell ke andar toh g zero hota hai, isliye wahan ye formula nahi chalta. Earth electricity ka conductor hai ya nahi, usse gravity ko fark nahi padta. Use bas itna chahiye ki mass poore volume mein barabar phaila ho.

Is lecture mein Earth uniform hai, tunnel smooth hai, hawa nahi hai, aur hum Earth ka spin ignore karte hain. Ab ek seedhi tunnel lo, jo centre se guzarti hui ek side ki surface se doosri side ki surface tak jaati hai. Surface wale end par mass m ki ball chhodo. Ye ball SHM karegi ya nahi, aur agar karegi, toh uska time period T kitna hoga?

Play from The centre is equilibrium; F = −kx

Pehle dekho ki ball kahan tiki reh sakti hai. Centre par g zero hai, toh ball par force bhi zero hai. Wahi uski equilibrium position hai, yaani SHM ki mean position. Ab ball ko centre se x door rakho, tunnel ke along. Right mein hum x ko plus maante hain, aur left mein minus maante hain.

Ball right mein centre se x door hai, toh r = x. Wahan g milta hai surface wale g × x/R. Isliye ball par pull ka size nikalta hai mgx/R.

Ye pull hamesha centre ki taraf hai, yaani left mein, x ke opposite. Ise hum F kehte hain. Uski unit newton hai, aur sign x se ulta hai.

Ab lecture 1 ki pehchaan yaad karo. SHM mein force likhte hain F = −kx, yaani mean position ki taraf aur displacement ke proportional. Yahan bhi F centre ki taraf hai aur x ke proportional hai. Toh ye wahi law hai, aur k banta hai mg/R.

Uski unit newton per meter hai, aur wo hamesha positive rehta hai. Baahar aisa nahi hota. Surface ke upar pull doori ke saath ghatta hai, proportional nahi rehta. Toh batao, ye law andar kitne x tak sahi hai?

Sahi. Centre se surface tak har r par exact hai g × r/R, toh force har x par exactly −kx rehta hai. Pendulum jaisa small angle wala approximation yahan nahi chahiye.

Play from T for any uniform planet, and the Earth

Isliye ball seedhi line mein aage-peeche jaati hai, yaani linear SHM karti hai. Ab hum lecture 4 ke do results lagate hain. Pehla result hai ω = √(k/m), aur doosra hai T = 2π/ω. Yahan k ki value rakho mg/R, toh k/m mein mass cancel ho jaata hai, aur bachta hai g/R.

Isliye ω banta hai √(g/R), aur T banta hai 2π√(R/g). Spring-block mein bhaari block dheema hota hai, par yahan pull bhi mass ke saath badhta hai. Isliye halki aur bhaari ball ka T ek hi hai.

Kisi planet ka g pata na ho, toh hum GM se kaam chalaate hain. Gravitation wale chapter mein G gravitational constant tha aur M Earth ka mass tha. Toh Earth ki surface par g kiske equal hota hai?

Sahi. Surface par g hota hai GM/R², yaani GM hota hai gR². Isi se g wala aur GM wala formula judte hain.

Ab g ki value rakho GM/R². Tab k, jo pehle tha mg/R, ban jaata hai GMm/R³. ω banta hai √(GM/R³), aur T banta hai 2π√(R³/GM). Ye kisi bhi uniform planet ke liye chalta hai, jiska mass M aur radius R ho. GM ki value rakho gR², toh T phir wahi nikalta hai, yaani 2π√(R/g).

Ab Earth ke numbers lete hain. Radius 6400 kilometer hai, yaani 6.4 × 10⁶ meter, aur g ki value 10 rakhte hain. R/g nikalo, toh 6.4 × 10⁵ aata hai, aur uska root hai 800 seconds. Ab T banta hai 2π × 800.

Ismein π ki value rakhein 3.14, toh 5024 seconds milte hain, yaani 83.7 minutes, kareeb 84 minutes. Ye value yaad rakhna. Lecture 16 ke bahut lambe pendulum mein bhi yahi formula aaya tha.

Play from The release point does not matter

Ab ball ko surface ki jagah aadhe raaste se chhodo, yaani centre se R/2 par. Use ab kam door jaana hai. Toh T ka kya hoga?

Sahi. Formula hai T = 2π√(R/g), aur ismein release point kahin nahi hai. Jab tak ball Earth ke andar hai, T wahi 84 minutes rehta hai. Amplitude aur sabse zyada speed zaroor badalte hain, yahan amplitude R/2 hai.

Surface se chhodne par amplitude A hota hai R, yaani centre se surface tak ki distance. Ball centre par sabse tez hoti hai. Dono surface wale ends par uski speed zero hoti hai, kyunki wahi turning points hain.

Play from The time to cross the tunnel

Ab time to cross dekhte hain. Lecture 2 mein humne padha tha ki A se mean position hote hue −A tak ka safar aadha oscillation hai. Toh ek side ki surface se doosri side ki surface tak ball kitne time mein pahunchegi?

Sahi. A se −A aadha oscillation hai, yaani T/2, kareeb 42 minutes. Surface se seedhe centre tak sirf chauthai safar hai, yaani T/4, kareeb 21 minutes.

Ab speed nikaalte hain, pehle energy se. Uske liye gravitation wale chapter ka potential chahiye. Earth ke andar, centre se r doori par, potential hota hai −GM × (3R² − r²)/2R³. Surface par, yaani r = R, ye banta hai −GM/R. Toh centre par, yaani r = 0, kitna hoga?

Sahi. r zero rakhne par bracket mein 3R² bachta hai, toh potential banta hai −GM × 3R²/2R³, yaani −3GM/2R. Ye surface se zyada negative hai, yaani centre par potential sabse kam hai.

Play from Speed at the centre: energy vs SHM

Potential energy ko hum likhte hain m × potential. Surface par ball ruki hui chhodi hai, toh wahan kinetic energy zero hai, aur potential energy hai −GMm/R. Centre par speed v maan lo. Tab kinetic energy hai (1/2)mv², aur potential energy hai −3GMm/2R.

Ab total energy dono jagah barabar rakhte hain, toh milta hai −GMm/R = (1/2)mv² − 3GMm/2R. Ise solve karo, toh milta hai (1/2)mv² = GMm/2R, yaani v = √(GM/R). GM ki value rakho gR², toh v banta hai √(gR).

Ab SHM wala raasta lete hain, jo chhota hai. Lecture 4 se hamein milta hai v max = ωA. Yahan ω hai √(g/R), aur A hai R, toh v max banta hai √(gR), yaani wahi answer.

Ab numbers rakhte hain. ω milta hai 1.25 × 10⁻³ radian per second, aur A hai 6.4 × 10⁶ meter. Tab v max hota hai 8000 meter per second, yaani 8 kilometer per second. Ulta sawal ho, yaani centre se ball ko itni speed do, toh wo surface tak pahunchkar hi rukegi.

Play from The speed at any r, and halfway

Centre ki jagah kisi bhi r par speed chahiye, toh lecture 4 ki speed lagao, yaani ω√(A² − x²). A ki jagah R aur x ka size r rakho. Tab speed banti hai √(g/R) × √(R² − r²).

Energy se bhi yahi nikalta hai. Usmein milta hai (1/2)mv² = GMm × (R² − r²)/2R³. Dono tareeke same hain, exam mein jo chhota lage wahi lagao.

Ab ek number lete hain. Surface se chhodi ball ki centre par speed 8 kilometer per second thi. Toh speed kitni hogi, jab r ho R/2?

Sahi. R/2 rakhne par root ke andar bachta hai R² − R²/4, yaani (3/4)R², aur uska root hai (√3/2) × R. Toh v banta hai (√3/2) × 8, yaani 0.866 × 8, kareeb 6.93 kilometer per second.

Play from A tunnel along a chord: restoring force

Ab tunnel centre se nahi guzarti. Earth mein ek seedhi smooth tunnel lo, jo centre se perpendicular doori D par hai, yaani ek chord. Uske surface wale end par ball rakhkar chhodo. Batao, iska T centre wali tunnel ke T se kaisa hoga?

Sahi. Ball wahi hai, toh uska mass nahi badalta, isliye T sirf k se tay hota hai, yaani T = 2π√(m/k). Toh ab hum chord par k nikaalte hain.

Pehle equilibrium position dhoondho, phir ball ko khiskao. Lecture 11 ke yahi steps hain, aur yahan bahut log pehla step chhod dete hain. Chord ke beech wale point par, jahan centre se aayi perpendicular tunnel se milti hai, pull theek centre ki taraf hai, yaani tunnel ke perpendicular. Use wall ka normal force N balance karta hai, aur tunnel ke along koi force nahi bachta. Wahi equilibrium position hai.

Ab ball ko equilibrium se right mein x door khiskao, tunnel ke along. Uski centre se doori r ho jaati hai, aur pull F ab bhi centre ki taraf hai, yaani tunnel ke perpendicular se tirchha. F aur tunnel ke perpendicular ke beech ke angle ko hum θ kahenge.

Centre par r aur D ke beech bhi yahi θ banta hai. Lecture 16 ki tarah F ke do components lete hain. F sin θ tunnel ke along hai, equilibrium ki taraf, aur F cos θ tunnel ke perpendicular hai, jise N balance karta hai.

Tunnel ke along net force banta hai −F sin θ, yaani equilibrium ki taraf. F ka size hai mgr/R, aur triangle se r sin θ = x. Toh F sin θ nikalta hai (mg/R) × x, aur net force banta hai −(mg/R) × x.

Ye −kx wala form hai. x ke saamne jo aata hai, wahi k effective hai, aur wo hota hai mg/R. Ye force ball ko equilibrium ki taraf lautata hai, isliye ise restoring force kehte hain.

Play from Any chord gives the same T

Yahan k wahi hai, toh ω bhi wahi hai, aur T banta hai 2π√(R/g), kareeb 84 minutes. D is formula mein hai hi nahi. Toh tunnel chahe centre se jaaye, yaani D zero, chahe D ho R/2, chahe surface ke paas ki chhoti chord ho, har tunnel mein T wahi rehta hai. Par amplitude badalta hai. Surface wale end se chhodein, toh amplitude hota hai √(R² − D²), jo D badhne par ghatta hai.

Time to cross bhi wahi rehta hai. Ek end se doosre end tak aadha oscillation hota hai, yaani T/2, kareeb 42 minutes, chahe chord lambi ho ya chhoti.

Ab wo chord lete hain, jismein D hota hai R/2. Surface wale end se chhodein, toh amplitude banta hai √(R² − R²/4), yaani (√3/2) × R, kareeb 0.866R. Beech wale point par speed sabse zyada hoti hai.

v max banta hai ωA, yaani 0.866 × ωR, jo 0.866 × 8 hai, yaani kareeb 6.93 kilometer per second. Yahi value centre wali tunnel mein bhi aayi thi, jab r tha R/2. Yaani surface se ruki hui chhodi ball ki speed bas centre se doori par depend karti hai.

Play from The wall's push on the chord, recap

Ab wall ke push N ko dekhte hain. N barabar hai F ke us hisse ke, jo tunnel ke perpendicular hai. Triangle se cos θ banta hai D/r, toh F cos θ nikalta hai (mgr/R) × (D/r), yaani mgD/R. Ismein x hai hi nahi, isliye N har jagah wahi rehta hai.

Ball beech mein aaye, toh θ zero ho jaata hai, aur left mein jaaye, toh θ ulti taraf banta hai, phir bhi N nahi badalta. Jab D hota hai R/2, tab N hota hai mg/2. Toh N versus x ka graph kaisa banega?

Sahi. N banta hai mgD/R, jismein x nahi hai. Toh graph x-axis ke parallel seedhi line hai, aur uski oonchai hai mg/2, kyunki D hota hai R/2. Jo x ke saath badhta hai, wo along wala hissa F sin θ hai.

Chalo recap karte hain. Earth ke andar tunnel ke along net force hota hai −(mg/R) × x, isliye ball SHM karti hai, centre wali tunnel mein bhi aur har chord mein bhi. T banta hai 2π√(R/g), kareeb 84 minutes, aur ek end se doosre end tak kareeb 42 minutes lagte hain. Centre wali tunnel mein surface se chhodein, toh centre par speed hoti hai √(gR), kareeb 8 kilometer per second. Agli lecture mein hum ring ki axis par particle dekhenge.

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