Consider a planet of radius \(r\) having density \(\lambda\).A tunnel is dig inside it it at a distance \(\frac{r}{2}\) from its centre as shown. An object of mass m is left in the tunnel at the surface at \(t=0\)

Find the time taken by the object to reach the mid of the tunnel and the normal reaction applied by wall of the tunnel on the object.

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TopNewestHint, it doesn't matter how far the tunnel is from the center of the Earth. – Josh Silverman Staff · 1 year, 11 months ago

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Can I have your views over here – Kushal Patankar · 1 year, 11 months ago

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First of all, thanks for mentioning me @Tanishq Varshney !! Is the answer for time period = \(\frac { \pi }{ 2 } \sqrt { \frac { 3 }{ 4\pi \lambda G } } \) ??? – Abhineet Nayyar · 1 year, 11 months ago

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– Kushal Patankar · 1 year, 11 months ago

I have same guess.Log in to reply

– Tanishq Varshney · 1 year, 11 months ago

Can u provide a hint or some expression to get the answerLog in to reply

Tunnel In a Planet

Let's consider that the object starts from the centre and moves towards the surface. Consider a very small displacement \('x'\) from the centre such that the restoring force remains constant. Now, the motion of the body through this \('x'\) is an SHM which i will prove later. According to shell theorem, the Earth around this motion won't apply any force. So, the force is applied by a small sphere with radius \(R=\sqrt { \frac { { r }^{ 2 } }{ 2 } +{ x }^{ 2 } } \) which is equal to \(\frac { Gm\frac { 4 }{ 3 } \pi \lambda { R }^{ 3 } }{ { R }^{ 2 } } \). But the restoring force will be the vertical component of this force = \(\frac { G(\frac { 4 }{ 3 } \pi { R }^{ 3 }\lambda )m }{ { R }^{ 2 } } \frac { x }{ R } \) = \(G\frac { 4 }{ 3 } \pi \lambda m(x)\)

Clearly this force is directly proportional to \(x\) and so, the motion is a simple harmonic one. So, the spring factor in the case is =\(G\frac { 4 }{ 3 } \pi \lambda m\) and the time period is given by = \(2\pi \sqrt { \frac { Mass\quad factor }{ Spring\quad factor } } \) = \(2\pi \sqrt { \frac { m }{ \frac { 4 }{ 3 } \pi \lambda Gm } } =2\pi \sqrt { \frac { 3 }{ 4\pi \lambda G } } \)

But since, you have asked about the time to move from extreme to mean, it is equal to \(Time period/4\) = \(\frac { \pi }{ 2 } \sqrt { \frac { 3 }{ 4\pi \lambda G } } \)

And the normal reaction (I guess :P) will be given by the other component of the attractive force = \(\frac { Gm\frac { 4 }{ 3 } \pi \lambda { R }^{ 3 } }{ { R }^{ 2 } } (\frac { r/2 }{ R } )\quad =\quad \frac { 2 }{ 3 } Gm\pi \lambda r\)

So, how did i do?? Lol:D – Abhineet Nayyar · 1 year, 11 months ago

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– Kushal Patankar · 1 year, 11 months ago

Thats it.Log in to reply

– Abhineet Nayyar · 1 year, 11 months ago

Yayy! lolLog in to reply

– Kushal Patankar · 1 year, 11 months ago

Since the motion inside is SHM , we know that force at amplitude is \(F=A \omega^2\) here A= √3r/2 and you know what F is. You will get \omega there. And the rest is piece of cake.Log in to reply

@Tanishq Varshney

Use dimensional analysis(during exam).

The actual way is to use restoring force and stuff, with it you will get that the motion of the object in the tunnel is S.H.M. – Raghav Vaidyanathan · 1 year, 11 months ago

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I need your view on this problem – Kushal Patankar · 1 year, 11 months ago

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@Raghav Vaidyanathan – Tanishq Varshney · 1 year, 11 months ago

I want that stuffLog in to reply

@Abhineet Nayyar @Kushal Patankar – Tanishq Varshney · 1 year, 11 months ago

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@Ronak Agarwal @Azhaghu Roopesh M @Rohit Shah @Krishna Sharma – Tanishq Varshney · 1 year, 11 months ago

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@Tanish – Ronak Agarwal · 1 year, 11 months ago

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– Prakhar Gupta · 1 year, 11 months ago

I totally messed up with the paper.Log in to reply

– Tanishq Varshney · 1 year, 11 months ago

The answer key webpage is showing error so i couldn't match my answersLog in to reply