Escape Velocity from Earth starting from its Centre

[Priyo137]

What is the minimum velocity (theoretically) of an object located at the CENTER of the Earth to move it outside its gravitational field?

 

[PeroK]

What is the minimum velocity (theoretically) of an object located at the CENTER of the Earth to move it outside its gravitational field?

Are you thinking about the Earth as a point mass or with a tunnel drilled theough to the centre?

 

[Priyo137]

I am looking for the second one, assuming a thin tunnel from the center to the surface. Further assuming that it does not necessarily change the distribution of mass of the spherical Earth severely.

 

[PeroK]

I am looking for the second one, assuming a thin tunnel from the center to the surface. Further assuming that it does not necessarily change the distribution of mass of the spherical Earth severely.

How much do you know about gravitational potential energy? What about Newton's Shell Theorem?

 

[Priyo137]

I have quite a basic idea of the two. It turned out to me that if we could know the energy required to bring the object to the surface of the Earth (by maybe using the concept of Gravitational potential energy), it can be solved.

 

[PeroK]

I have quite a basic idea of the two. It turned out to me that if we could know the energy required to bring the object to the surface of the Earth (by maybe using the concept of Gravitational potential energy), it can be solved.

That's exactly how you would do it. I'll have a look on line for a good derivation, rather than typing lots in here!

This calculation is sometimes carried out for the hypothetical gravity train:

https://en.wikipedia.org/wiki/Gravity_train

 

[Orodruin]

Since the Earth density profile is not a simple analytic function, this computation has to be performed numerically. However, given the density profile data, it should be a simple matter of numerical integration to find the gravitational potential at the centering.

 

[PeroK]

Since the Earth density profile is not a simple analytic function, this computation has to be performed numerically. However, given the density profile data, it should be a simple matter of numerical integration to find the gravitational potential at the centering.

At a B level we have to assume constant density.

 

[PeroK]

That's exactly how you would do it. I'll have a look on line for a good derivation, rather than typing lots in here!

I couldn't find anything very good, so might post something myself later.

 

[Orodruin]

At a B level we have to assume constant density.

Unfortunately, that will tend to significantly underestimate the gravitational potential at the center. You will get an order of magnitude estimate, but that you could get from just dimensional analysis.

Arguably, B level might also rule out the use of the shell theorem. If we allow integrals, then a simple numerical approach based on existing data is not much more advanced.

 

[PeroK]

I couldn't find anything very good, so might post something myself later.

First, we start with Newton's law of gravitation:
$$F = \frac{Gm_1m_2}{r^2}$$This gives the magnitude of the gravitational force ($F$) between two point masses ($m_1$ and $m_2$) that are a distance $r$ apart. The force is attractive, of course. For two objects we can often use $r$ as the distance between the centres of mass to get a good approximation of the gravitational force between them. In fact, if the objects are spherically symmetric, then we can take $r$ as the distance between the centres and this works out exactly. This is known as Newton's shell theorem, which requires some significant calculus to prove. See here, for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html#wtls

Moreover, the gravitational force on an object inside a spherical shell is zero(!) This can also be proved using calculus:

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html

The conclusion is that for an object of mass $m$ outside the Earth (of mass $M$), the gravitational force on the object is given by:
$$F = \frac{GMm}{r^2}$$Where $r$ is the distance to the centre of the Earth. And, for an object inside the Earth, the gravitational force is due only to the spherical mass closer to the Earth's centre. The total gravitational force of the Earth's mass further from the centre than the object is zero. If the radius of the Earth is $R$, then the gravitational force on an object a distance $r < R$ from the centre of the Earth is:$$F = \frac{GM(r)m}{r^2}$$Where $M(r)$ is the mass of the Earth up to the radius $r$.

To make further progress, we assume that the Earth is of uniform density. In that case we have:
$$M(r) = \frac{r^3}{R^3}M$$I'll leave it as an exercise to prove that.

This gives us the force on an object at radius $r$ inside the Earth as:
$$F = \frac{GMmr^3}{R^3r^2} = \frac{GMmr}{R^3}$$That is a varying gravitational force, so to get the speed of an object falling from the surface to the centre, we would need to use more calculus (integration). One way to do this is to calculate the gravitational potential energy inside the Earth. This turns out to be:
$$U(r) = \frac{GMmr^2}{2R^3}$$Note that the potential energy at the centre is zero: $U(r=0) = 0$. And, the potential energy at the surface is:
$$U(r=R) = \frac{GMmR^2}{2R^3} = \frac{GMm}{2R}$$
This tells us the kinetic energy that the object gains when falling from the surface to the centre, or loses when moving out from the centre to the surface.

To get the escape velocity from the centre, we also need the escape velocity from the surface. The gravitational potential energy gained by moving from the Earth's surface to a long way away ("infinity") is:
$$\Delta U = \frac{GMm}{R}$$Where again $R$ is the radius of the Earth.

We can put these two together to get the kinetic energy needed to escape from the centre of Earth to "infinity":
$$KE = \frac{GMm}{2R} + \frac{GMm}{R} = \frac{3GMm}{2R}$$And, from this you can calculate the escape velocity. I'll leave that as an exercise.

Note that as @Orodruin has pointed out, the Earth does not have uniform density, so it would be an interesting exercise to do a numerical calculation for the real Earth, rather than the hypothetical earth of uniform density.

 

[berkeman]

Welcome to PF, @Priyo137

What is the minimum velocity (theoretically) of an object located at the CENTER of the Earth to move it outside its gravitational field?

Is this question for a school assignment, or for your general interest?

 

[Priyo137]

Welcome to PF, @Priyo137


Is this question for a school assignment, or for your general interest?

Thanks for welcoming me here!
It's a question from a very recent physics exam at my high school.

 

[berkeman]

It's a question from a very recent physics exam at my high school.

Okay, then I'll move your thread to the schoolwork forums. Also, for schoolwork questions, you need to show us your work on the problem. Can you do that using the hints that you've been provided?

(I'll send you some tips on learning to post math using LaTeX at PF.)

 

[PeroK]

Take a look at the variation in gravitational acceleration inside the Earth here:

https://en.wikipedia.org/wiki/Gravity_of_Earth

The acceleration is approximately constant ($10m/s^2$) from the surface to a radius of about $3,000km$ and reduces linearly after that. This provides the option of a simple model based on two phases for the acceleration Earth. The first phase should be straightforward and the second phase can use the same method as above.

This gives a more accurate estimate for the escape velocity from the centre (for what it's worth!). Here's the graph from the Wikipedia page:

 

[Orodruin]

Take a look at the variation in gravitational acceleration inside the Earth here:

https://en.wikipedia.org/wiki/Gravity_of_Earth

The acceleration is approximately constant ($10m/s^2$) from the surface to a radius of about $3,000km$ and reduces linearly after that. This provides the option of a simple model based on two phases for the acceleration Earth. The first phase should be straightforward and the second phase can use the same method as above.

This gives a more accurate estimate for the escape velocity from the centre (for what it's worth!). Here's the graph from the Wikipedia page:

View attachment 342053

Just to note that the gravitational potential at the center is the area under the curve here. The difference between PREM and constant density is appreciable. The area outside of the Earth radius is easy to underestimate, but it is the same for all models obviously.

 

[PeroK]

Just to note that the gravitational potential at the center is the area under the curve here. The difference between PREM and constant density is appreciable.

I make it about 20%.

 

[Priyo137]

Okay, then I'll move your thread to the schoolwork forums. Also, for schoolwork questions, you need to show us your work on the problem. Can you do that using the hints that you've been provided?

(I'll send you some tips on learning to post math using LaTeX at PF.)

Yes, sure.
We have already got,
$$KE=\frac{3GMm}{2R}$$
Let the escape velocity from the centre be $v_{ec}$, so we can write,
$$\frac{1}{2}mv^2_{ec} = \frac{3GMm}{2R}$$
Through algebraic manipulations and substituting the value of $g = \frac{GM}{R^2}$ at Earth's surface (just to make calculation and typing easier!), we get,
$$v_{ec}= \sqrt{3gR} = 13.717 km/s$$
Which is a plausible velocity, intuitively higher than the escape velocity from the Earth's surface (11.2 km/s).
I hope it's correct. If not, please let me know.
Thank you everyone.

 

[Orodruin]

$$v_{ec}= \sqrt{3gR} = 13.717 km/s$$

A comment regarding the precision here. A fundamental skill in physics (unfortunately many times ignored) is to not over or understate the precision in your results. Whether the precision is limited by limited precision in thr input data or by inaccuracies in the modelling. How many significant digits would you be confident with in this result?

 

[Orodruin]

Also, here is a very basic Python implementation using a sum approximation of the appropriate integral:

import numpy as np
import pandas as pd

data = pd.read_csv('PREM_1s.csv',header=None)
rho = data[2] * 10**-3 * (10**2)**3
r = data[0] * 10**3
dr = - r.diff()
G = 6.674 * 10**-11
print('Escape velocity at center: ' + str(np.sqrt(2*(rho * 4 * np.pi * r * G * dr).sum())) + ' m/s')

The output is ca 14.8 km/s. I still would not be comfortable with quoting more digits than that.

Note: So this means a relative difference of $14.8/13.7 - 1 \simeq 0.08$ in the escape velocity, corresponding to a relative difference of about $1.08^2 - 1 \simeq 0.17$ in the potential so this

I make it about 20%.

is a pretty decent guess.

Data csv-file used is taken from the Preliminary Reference Earth Model here.

 

[Priyo137]

A comment regarding the precision here. A fundamental skill in physics (unfortunately many times ignored) is to not over or understate the precision in your results. Whether the precision is limited by limited precision in thr input data or by inaccuracies in the modelling. How many significant digits would you be confident with in this result?

Unfortunately, it's hard for me to approximate. All I can guess that it's going to significantly change if we're taking real data of the account. The major change will be causes probably for the assumption that Earth has a constant density.
So you see, I can't exactly insist on how many significant digits it's going to be precise.

 

[Orodruin]

Unfortunately, it's hard for me to approximate. All I can guess that it's going to significantly change if we're taking real data of the account. The major change will be causes probably for the assumption that Earth has a constant density.
So you see, I can't exactly insist on how many significant digits it's going to be precise.

Then you should train this skill. You have already seen @PeroK estimate the error in the potential to about 20%. Does it seem reasonable to then use five significant digits in your result?

 

[Priyo137]

Then you should train this skill. You have already seen @PeroK estimate the error in the potential to about 20%. Does it seem reasonable to then use five significant digits in your result?

I assume I shouldn't have used 5 significant digits. Maybe I should have kept 1 only, I'm not sure. It would be great if you helped. Thanks in advance.

 

[Orodruin]

I assume I shouldn't have used 5 significant digits. Maybe I should have kept 1 only, I'm not sure. It would be great if you helped. Thanks in advance.

If you know that there is around a 10% error (which is roughly what you get if you take the square root of a number with a 20% error) then you should keep a number of digits reflecting this. Either one or two in this case, neither is ideal as one gives you the order of magnitude only and two kind of oversells precision in the end (14 km/s when correct rounding of the actual result would be 15 km/s). Of the two, I would probably go for two in this case but make sure to note that the second digit might not be fully accurate.

Using five gives an impression of precision that simply is not there.

 

[PeroK]

is a pretty decent guess.

It wasn't a guess. I calculated the two-phase model on a spreadsheet and got $14.9 km/s$.

 

[PeroK]

Unfortunately, it's hard for me to approximate. All I can guess that it's going to significantly change if we're taking real data of the account. The major change will be causes probably for the assumption that Earth has a constant density.
So you see, I can't exactly insist on how many significant digits it's going to be precise.

That's the assumption that makes the solution unrealistic, and not a good approximation. It's a challenge to find a method of getting a better approximation from the data. That involves really thinking about the problem and how you are going to solve it, as well as using equations and doing calculations.

If you can't program in Python, you can do a lot on a spreadsheet. In general, for maths and physics problems I find a spreadsheet a lot better than a calculator.

 

[Orodruin]

If you can't program in Python,

… then you should learn to program in Python.

Tounge in cheek of course, but it is actually really intuitive, easy to use, has a lot of features and packages, and allows you to do a lot as long as you can find proper data for input. And I am saying this as someone who jumped on the Python train just this past fall. It is now my go-to for quick numerical computations after using Matlab for that for essentially all of my carreer.

you can do a lot on a spreadsheet. In general, for maths and physics problems I find a spreadsheet a lot better than a calculator.

All have their uses. Spreadsheets are one step up from a calculator in complexity abd you can (if you want) program them to add extra functionality. Scripting languages are a step up from that and compiled languages a step up from that. All tools have their uses. Some are better than others for particular purposes.

 

[PeroK]

I assume I shouldn't have used 5 significant digits. Maybe I should have kept 1 only, I'm not sure. It would be great if you helped. Thanks in advance.

There are various ways of deciding how many significant digits to keep. The data I was using had three significant digits, so I would tend to keep only three in the answer. One way to check this is to use values either side of the data to see how much that changes things. I did this on a spreadsheet. The calculation for the difference in GPE was:
$$\Delta U = \frac{3GM}{2R}$$I did three calculations: the first with the data as it was. The second with a slightly lower value for $G$ and $M$ and a slightly greater value for $R$. This should give a lower answer. The third with a higher value for $G$ and $M$ and a lower value for $R$. This should give a greater answer. Then we can see how much difference these small changes make. I put this in a spreadsheet:

		Min	Max
G	6.67E-11​	6.66E-11​	6.68E-11​
R	6.37E+06​	6.36E+06​	6.38E+06​
M	5.97E+24​	5.96E+24​	5.98E+24​
Delta U	9.38E+07​		Best estimate
V	13,694​
Delta U	9.33E+07​		Minimum estimate
V	13,662​
Delta U	9.42E+07​		Maximum estimate
V	13,727​

You can see that the estimates ranged from $13.662 km/s$ to $13.727 km/s$, with a best estimate of $13.694 km/s$. This suggests that $13.7 km/s$ is appropriate and not to use any more significant figures.