Energy Released From Dropping Things Onto Neutron Stars

[RyanJ]

TL;DR Summary

Calculating the energy released when dropping things onto neutron stars.

Hi all.

I'm trying to get my head back onto mathematical thinking and I'm setting myself projects to undertake to facilitate that. It's been too long since I've worked with numbers like this. My first task was to work out how much energy would be released when dropping things onto a neutron star. I've often heard things along the lines of "dropping something from a meter onto the surface of a neutron star would release more energy than every nuclear weapon detonated on the Earth" - and I wanted to validate that claim for myself.

I've run the numbers and I'm certain there is something wrong since the numbers come to nowhere near that. I've just accounted for the kinetic energy here, I've got no idea where to start with other energy release factors - please do elaborate on those. I'm aware that this is a longer method and that there are shorter ones, but I wanted to get in the most practice I could here.

The Constants

G = $6.6710^{-11}$
Mass of object (m) = $0.0459 kg$
Mass of neutron star (M) = $4.672910^{30} kg$ (~2.4 solar masses)
Radius of neutron star (r) = $15000$ m
Height above surface (h) = $1$ m

Step 1 - calculate the acceleration due to gravity

$$g = \frac{GM}{r^{2}}$$
$$g = \frac{6.6710^{-11} × 4.672910^{30}}{5000^2}$$
$$g = \frac{3.116824310^{20}}{2.2510^8}$$
$$g = ~1.3852610^{12}$$
$$g = 1.3910^{12}$$

Step 2 - calculate the time taken for the object to fall over the distance h

$$t = \sqrt{\frac{2h}{g}}$$
$$t = \sqrt{\frac{2 × 1}{1.3910^{12}}}$$
$$t = \sqrt{1.4388510^{-12}}$$
$$t = 1.1995210^{-6} s$$

Step 3 - calculate the velocity of the mass falling at an acceleration of g for a time t

$$v = v_0 + gt$$

But, since the object has no initial velocity it simplifies to...

$$v = gt$$
$$v = 1.3910^{12} × 1.1995210^{-6}$$
$$v = 1.33733 × 10^6 ms^{−2}$$

Step 4 - calculate the kinetic energy of the mass at the end of the fall

Note, I'm using the relativistic kinetic energy equation here, since I'm concerned that the speeds could start to fall within the realm where relativity starts to become significant to the end result.

$$k_e = m_0 × c^2 × (\sqrt{1 - \frac{v^2}{c^2}} - 1)$$
$$k_e = 0.0459 × 299792458^2 × (\sqrt{1 - \frac{(1.33733 × 10^6)^2}{299792458^2}} - 1)$$
$$k_e = 0.0459 × 89875517873681764 × (\sqrt{1 - \frac{1788451528900}{89875517873681764}} - 1)$$
$$k_e = 0.0459 × 89875517873681764 × (\sqrt{1 - 1.9899e-5} - 1)$$
$$k_e = 0.0459 × 89875517873681764 × (\sqrt{0.999980101} - 1)$$
$$k_e = 0.0459 × 89875517873681764 × (0.999990050 - 1)$$
$$k_e = 0.0459 × 89875517873681764 × -0.00000995$$
$$k_e = ~-41046598390.5 J$$

So, that's around 41 GJ - which is significantly lower than the yield of a nuclear blast, let alone all of the nuclear bombs ever detonated. From memory a TNT equivalent for 1 kiloton is around 4.2 TJ - meaning it is somewhere around 0.01 kilotons.

Have I made a mistake in my calculations or are there other factors I need to include too? I'm doing this largely from memory, so please don't hesitate to point out any mistakes, assumptions or oversights in my calculations. I need the practice, and that's not going to do any good if the results are incorrect.

Edit: fixed the equation formatting.

 

[mfb]

This is much easier to calculate using the potential energy E = GMm/r: The released energy is $\Delta E = GMm (\frac 1 {r_2} - \frac 1 {r_1})$ where $r_1 = 15001 m$ and $r_2 = 15000 m$. For you values this is 63 GJ.

Your made a multiplication error where you calculated the velocity, fix that and you should get the same result.

The object is below 1% the speed of light so you don't need to use the relativistic formula - it's more likely to make the result less accurate because it's very sensitive to rounding errors. Assuming a constant acceleration over that meter is a larger approximation than Newtonian mechanics here.

I've often heard things along the lines of "dropping something from a meter onto the surface of a neutron star would release more energy than every nuclear weapon detonated on the Earth"

That obviously depends on the mass of the object. If the statement doesn't include the mass, then you know it shouldn't be taken very seriously.

 

[Vanadium 50]

I've often heard

Where?

Also, I suggest in the future you work algebraically as long as possible - its easier to see what you are doing - rather than a wall of numbers.

 

[phinds]

I've often heard

I see you're new, so I assume you're not aware that as far as PF is concerned, that has exactly as much validity as "some guy on the subway told me that ... ".

Better here on PF to just ask your question without trying to back-fill your rationale with a useless "citation".

 

[RyanJ]

Also, I suggest in the future you work algebraically as long as possible - its easier to see what you are doing - rather than a wall of numbers.

Off hand, I remember this one from several years ago. I had to go back and look it up. Yes, I am aware YouTube isn't necessarily always a source of reliable information.



I also recall hearing a similar claim from documentaries in the past, one I specifically remember was this one (the clip is of awful quality). This one was made by the BBC. There are definitely others, but it would take some effort to dig them all up.



I'm not satisfied with the whole "because it was said" argument, so I figured I'd try to calculate it myself and find out. Though I did doubt my calculation due to the disparity between the results and the stated results.

Also, I suggest in the future you work algebraically as long as possible - its easier to see what you are doing - rather than a wall of numbers.

Thanks for the suggestion. I'll try to do that in the future!

 

[sophiecentaur]

TL;DR Summary: Calculating the energy released when dropping things onto neutron stars.

Step 3 - calculate the velocity of the mass falling at an acceleration of g for a time t

You are making the assumption that the acceleration is uniform over the time of the fall. Obvs it works for a 1m drop on Earth but you might calculate the difference in acceleration when dropping onto a neutron star. That could be revealing.
@mfb 's formula, based on the change in potential is the way to go. It's both easier and more accurate.

The term 'spaghettification' is applied to the likely effect on objects falling onto black holes. This is due to the high gradient of gravitational potential and you could use the results of your g calculations at different heights to estimate the stretch on your falling object (of finite size). These calculations are very easy and it's only because I am a lazy houndog that I haven't done them myself. (an exercise for the student, lol)

 

[RyanJ]

The term 'spaghettification' is applied to the likely effect on objects falling onto black holes. This is due to the high gradient of gravitational potential and you could use the results of your g calculations at different heights to estimate the stretch on your falling object (of finite size). These calculations are very easy and it's only because I am a lazy houndog that I haven't done them myself. (an exercise for the student, lol)

Hmm. Let's see. I believe we can use the following equation for this calculation.

$F = \frac{GmM}{(r+h)^2}$

If we assume that the golf ball is a standard one, then it's height is equal to its diameter (d), so 0.04267 meters. If the bottom of the ball is at 1 meter above the surface (h) then the top will be at the h plus d, giving h_top.

$F_{top} = \frac{GmM}{(r+h_{top})^2}$
$F_{bottom} = \frac{GmM}{(r+h)^2}$

Solving this for a neutron star with the specifications in my original post yields the following:

$F_{top} = 6.35745×10^{10} N$
$F_{bottom} = 6.35749×10^{10} N$

That gives the $\Delta F$ as being ~400000 Newtons over the height of the ball. That's an insane number.

 

[A.T.]

The NASA guy at 0:43 talks about dropping a marshmallow on a neutron star being equivlant to an atomic bomb, but he doesn't specify the drop height. Maybe he means from far away in space.

 

[willem2]

That gives the ΔF as being ~400000 Newtons over the height of the ball. That's an insane number.

Yet, the golf ball won't deform very little, during the time it falls. The acceleration of the tidal force is only 400000/0.045 = (0.045 kg golf ball) = 8.9 * 10 ⁶ ms ^-2
(1/2)at² = 0.01 mm.

 

[sophiecentaur]

That's an insane number.

It certainly is. But don't you feel better, having proved it?
I should have done it in my head using a 10,000:1 ratio of distances. Plus you square the ratio for the force. I'd imagine a large volume (sphere) of gas (say 1km diameter) would stretch noticeably on the way down.

 

[RyanJ]

Yet, the golf ball won't deform very little, during the time it falls. The acceleration of the tidal force is only 400000/0.045 = (0.045 kg golf ball) = 8.9 * 10 ⁶ ms ^-2
(1/2)at² = 0.01 mm.

Which is also a crazy thought. I would have thought the force would have been more than enough to "make the golf ball dead" at that distance.

Either way, the poor golf ball would be in for a very bad time. The surface of the neutron star, as I understand it, is all but indestructible as far as "normal" matter is concerned. Splat.

 

[sophiecentaur]

Either way, the poor golf ball would be in for a very bad time.

Scratch the surface of this question and we're in the territory of "What would happen if the Sun suddenly ceased to exist?" The first (PF style) reaction to that one is 'How would you make that happen?' and we really should perhaps ask how the proposed neutron star experiment could be done.
The gpe of the 1kg mass would be around 10¹⁵J, near the surface, so to arrange for the mass to start off stationary, you'd need a 're-entry' rocket to provide this much energy to brake the mass's fall. That's around 2,000TJ which is a typical yield from a big fission bomb. See the wiki page.

 

[A.T.]

The gpe of the 1kg mass would be around 10¹⁵J, near the surface, so to arrange for the mass to start off stationary, you'd need a 're-entry' rocket to provide this much energy to brake the mass's fall. That's around 2,000TJ which is a typical yield from a big fission bomb. See the wiki page.

Likely that's the equivalence to atomic bombs that those pop-sci sources are talking about.

 

[jbriggs444]

You are making the assumption that the acceleration is uniform over the time of the fall. Obvs it works for a 1m drop on Earth but you might calculate the difference in acceleration when dropping onto a neutron star. That could be revealing.
@mfb 's formula, based on the change in potential is the way to go. It's both easier and more accurate.

More accurate, certainly. But if we are looking for one sig fig in the result, a 4 sig fig answer will do just fine. And without the numeric ill-conditioning involved with subtracting two nearly equal figures. [I learned with 32 bit floating point and try to stay away from ill conditioned formulae].

$\Delta PE = mgh$ is pretty simple.

Let us work the numbers.

We have @mfb suggesting:
$$\Delta PE = GMm(1/r_2-1/r_1)$$where $r_1 = 15001 \text{ m}$ and $r2 = 15000 \text{ m}$

We have the competing formulae depending on where we take $g$ to be defined:
$$\Delta PE = GMm\frac{r_1 - r_2}{{r_1}^2}$$ $$\Delta PE = GMm\frac{r_1 - r_2}{{r_2}^2}$$for the same values of $r_1$ and $r_2$.

Let us evaluate those three right hand terms:
$$\frac{1}{15001} - \frac{1}{15000} \approx -4.444 \times 10^{-9}$$
$$\frac{15000-15001}{15001^2} \approx -4.443 \times 10^{-9}$$
$$\frac{15000-15001}{15000^2} \approx -4.444 \times 10^{-9}$$

 

[sophiecentaur]

I was scanning through the wiki page about neutron stars to follow up the implied consequences of all those numbers. It makes good reading and I don't doubt that most of Wiki's numbers are about right.

Neutron stars are (as you might expect) sources of high energy Xrays as the consequence of accretion - one fission bomb's worth for each collision. I was wondering about the likely amount of accretion you could expect for a neutron star at the centre of a load of products of the original collapse. The mechanism for accretion is the interaction of the various bands in the accretion exchanging momentum and producing 'slow' fragments which fall straight in. Also, there are images of neutron stars stealing matter from neighbours. Neutron stars are very bright and (so they say) the majority of the radiated energy is from accretion, rather than nuclear processes inside energy is due to accretion (interesting or perhaps obvs for some readers)

I did find a confusing statement in that link, though (End of the 'formation' section):
" The neutron star's density also gives it very high surface gravity, with typical values ranging from 1012 to 1013 m/s2 (more than 1011 times that of Earth).[21] One measure of such immense gravity is the fact that neutron stars have an escape velocity of over half the speed of light.[22] The neutron star's gravity accelerates infalling matter to tremendous speed, and tidal forces near the surface can cause spaghettification.[22]"
It agrees with our established value of surface gravity but their value for escape velocity seems to be very different. I can't reconcile the two. How are they calculating escape velocity, if not by equating gpe with KE? Could it be by assuming constant acceleration? That's dodgy.

 

[mfb]

It agrees with our established value of surface gravity but their value for escape velocity seems to be very different.

Different from what? I don't see anyone else calculating an escape velocity here. OP's values and Newtonian gravity would predict an escape velocity of 206,000 km/s or 2/3 the speed of light, which is on the high side because OP's mass value is very high for a neutron star and I didn't consider relativistic effects.

If you assume constant acceleration then there is no escape velocity.

 

[sophiecentaur]

Different from what? I don't see anyone else calculating an escape velocity here.

I did find a confusing statement in that link

You just told me I was wrong which doesn't surprise me. I had said I was confused and was asking for some help with why.

If escape velocity is √(2gpe), that seems to come to 1.4X10⁷ which is very sub c. They describe the situation as causing spaghettification.

 

[jbriggs444]

You just told me I was wrong which doesn't surprise me. I had said I was confused and was asking for some help with why.

If escape velocity is √(2gpe), that seems to come to 1.4X10⁷ which is very sub c. They describe the situation as causing spaghettification.

Spaghettification relates to the gradient of gravitational acceleration.
Escape velocity relates to the integral of gravitational acceleration.

Yes, spaghettification is a concern. If your gravitational acceleration is $10^{12}$ g then changing from the inverse square of 15001 to the inverse square of 15000 is only a .013 percent increase but 0.013 percent of $10^{12}$ is still over $10^8$ g's.

Not enough to change the PE calculation by much but plenty to snap a cable with.

 

[mfb]

If escape velocity is √(2gpe), that seems to come to 1.4X10⁷ which is very sub c. They describe the situation as causing spaghettification.

The gravitational potential energy people calculate here is for 1 meter above the surface relative to the surface. For the escape velocity you need the energy relative to infinity.

 

[sophiecentaur]

Not enough to change the PE calculation by much but plenty to snap a cable with.

Right. The gradient difference between inverse and inverse square. Makes sense now. Thanks.

 

[jbriggs444]

Right. The gradient difference between inverse and inverse square. Makes sense now. Thanks.

The gradient (of the acceleration) is inverse cube. Mathematically, this is a tensor field.
The acceleration is inverse square. Mathematically, this is a vector field.
The potential is simple inverse. Mathematically, this is a scalar field.