Gravitational Force and Density on a Planet with a Digging Experiment

[Raghav Gupta]

Homework Statement

A planet of radius R = (radius of earth)/10 has the same mass density as earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and of linear mass density 10^-3 kgm^-1 into it. If the wire is not touching anywhere , the force applied at the top of the wire by a person holding it in place is ( take the radius of Earth = 6 x 10⁶m and the acceleration due to gravity on Earth is 10 ms^-2 )
96
108
120
150

Homework Equations

I think there is some formula for gravitational acceleration when we dig a depth on planet.
Don't know that or the derivation.

The Attempt at a Solution

Density = mass/volume
Volume = 4/3πR³
I do not know what is happening by digging.

 

[BvU]

Hello Raghav,

I think there is some formula for gravitational acceleration when we dig a depth on planet

Your intuition serves you well. There certainly is.


There is a lot of good stuff to be found under "electric field of uniformly charged non-conducting sphere" (and the next part, same page) because gravitational field and electric field are -- mathematically speaking -- the same.

You have to understand Gauss's theorem and then it boils down to: "below the surface you only have to consider what's lower than yourself to calculate graviational force". All the way down to the center of the sphere, where the force is zero.

You end up with an integration of the varying gravitational force along the length of the wire.

--

 

[Raghav Gupta]

Hello Raghav,Your intuition serves you well. There certainly is.There is a lot of good stuff to be found under "electric field of uniformly charged non-conducting sphere" (and the next part, same page) because gravitational field and electric field are -- mathematically speaking -- the same.

You have to understand Gauss's theorem and then it boils down to: "below the surface you only have to consider what's lower than yourself to calculate graviational force". All the way down to the center of the sphere, where the force is zero.

You end up with an integration of the varying gravitational force along the length of the wire.

--

So from the links you have given, $g = \frac{GM\frac{R}{5}}{\frac{R^3}{10^3}}$ ?

 

[BvU]

I take it that with R you mean the small planet radius, as in the problem statement ?
And with g you mean the gravitational acceleration at some distance from the center of this planet ? What distance ?
Why the $R^3/10^3$ ?

 

[Raghav Gupta]

I take it that with R you mean the small planet radius, as in the problem statement ?
And with g you mean the gravitational acceleration at some distance from the center of this planet ? What distance ?
Why the $R^3/10^3$ ?

Sorry, I have directly put the values given in question in formula.
I should properly understand.
Suppose I have in the relevant equation tool now,
F = Gm₁m₂/r² where m1 and m2 are masses of body and r the distance between them.
Can I solve the problem, with this much information?
I also know high school level integration.

 

[Raghav Gupta]

I was trying some things, is this correct?
Force on small part of wire is G m_planet dm/x² , now dm = Mdx/L where L is length of wire.
L = R/5
So ,
$$ F = \int_{4R/5} ^R \frac{G m_{planet} dm}{x^2} $$
dm = Mdx/L ?

 

[nasu]

No, if by m_planet you mean the mass of the entire planet (a constant).
Why don't you start by finding an expression for the gravitational acceleration as a function of position in the well?

 

[Raghav Gupta]

No, if by m_planet you mean the mass of the entire planet (a constant).
Why don't you start by finding an expression for the gravitational acceleration as a function of position in the well?

Hey, what if I write here m_planet as ρx 4/3 x πx³ ? Where ρ is density of planet? Density we can consider here same.
Then we can easily solve this.

 

[BvU]

Notation is a bit confusing, but you if you mean $m(x) = \rho \;\tfrac {4} {3} \pi x^3$ I really like it. Just don't call it m_planet...

So what does this mean for the gravitational acceleration as a function of the distance to the center of the planet ?

 

[Raghav Gupta]

I think as we have to find force,
F =

$$ F = \int_{4R/5} ^R \frac{G ρ\frac{4πx^3}{3} dm}{x^2} $$
dm = Mdx/L , where M/L is linear mass density of wire = 10^-3
On some solving,
$$ F =\frac{Gρ4π10^{-3}}{3} \int_{4R/5} ^R x dx $$
Write ρ = m_earth / Volume of earth
as ρ is same for both planet.
Write G m_earth = g radius²_earth, here g = 10
On solving and solving, I
got the answer F = 108 N

 

[BvU]

Excellent work !

 

[AdityaDev]

Jee Advanced physics 2014. Its a good question. But not difficult.

 

[BvU]

Quite a cottage industry, this JEE business. Any link for the exercise texts (without joining and all that stuff) ?

 

[Raghav Gupta]

Quite a cottage industry, this JEE business. Any link for the exercise texts (without joining and all that stuff) ?

I would be glad if we can discuss some of the hard seeming questions of physics and Maths in this paper ( I have not seen you in chemistry section).
Here is the
http://jeeadv.iitb.ac.in/sites/default/files/2014p2key.pdf .

 

[AdityaDev]

How am I supposed to remember colours of all inorganic compounds? Look at numerical question in advanced chemistry paper 1. It's available on the same site you specified.

 

[Raghav Gupta]

How am I supposed to remember colours of all inorganic compounds? Look at numerical question in advanced chemistry paper 1. It's available on the same site you specified.

Don't know. It's a problem for me as well.
Try suggesting that topic for PF Insights in STEM forum.