Challenge Questions

First step: Equate PE of lava (at max height) to KE (at surface).

½mv² = mg_ph => g_p = ½v²/h = ½ * 150 * 150 / 4000 = 2.8125ms^-2

m = mass of lava (which cancels out)

g_p = acceleration due to gravity on planet.

So (ignoring the planetary radius), mass of planet M_p ≈ M_earth * 2.8 / 9.8 ≈ 6e24 * 0.287 ≈ 1.72e24 kg

(g_earth = 9.8 ms^-2)

Getting late - someone else can take over ...

Hi, Jorrie,

I realized my "answer" wasn't correct - as I said I hadn't taken the relative radii into account. If it hadn't been 1am on a Sunday night/Monday morning, I'd like to think I would have got round to using the square of the ratio of radii as a "correction factor".

Thanks for saving me the Monday Morning Brain Strain .

I was perplexed by the discrepancy (in the last decimal place) between your answer an AH's.

You obviously know the Gravitational constant "off by heart" as 6.67, which gives an answer of 2.6354, and you have rounded up to 2.64.

Using the value in Wikipedia gives a range of answers from 2.62298 to 2.63345, so accepting that AH was curiously out by a factor of 10,000 instead of 1 million his last digit was probably more accurate.

So what? I'm just curious: Wikipedia actually has :-

Do we really not know G more accurately than that?

Hi Randal, you asked: "Wikipedia actually has :-

Do we really not know G more accurately than that?"

I would think that's pretty accurate! In the case of Earth-orbiting satellites, other uncertainties, like the inhomogeneities in the gravitating mass and small discrepancies in the height of the satellite have bigger influences than that.

-J

I thought that G was almost irrelevant for most satellites, as we try to space them so that only the "G x mass" product of the local planetary bodies is important. But mightn't an experiment where we have a gravitationally-coupled pair of high-orbit satellites be capable of improving on the present situation? (Hard to see who would pay for this at the present, perhaps)

With "try to space them so that only the "G x mass" product of the local planetary bodies is important", are you referring to gravity between the satellites that is irrelevant?

My point was that despite possible errors in G that make our position predictions for individual satellites less than perfect, it is dwarfed by the many other effects, e.g. varying density of the Earth underneath them, solar wind, etc.

I suppose a gravitationally-coupled pair of satellites (as you said) in orbit around the Sun could perhaps improve the value of G, but, if already good enough, why bother?

-J

As you say, this ignores atmosphere, rotation*, non-uniformity of planet and of eruption, influence of moons, and probably quite a few other things.

However, the reason for my comment is that, at least in principle, I would prefer to use the differences in potential energy (proportional to mass/radial distance) rather than height x acceleration. But it really doesn't matter here, as it only makes about 0.16% difference.

*I would have thought that volcanoes would be unlikely without something to generate tidal forces, but I could be wrong.

Not my area of experience.

Can't read the equations.

I'm interested though.

Curious what the Universal Gravity Equation means really. Does that mean periodic table elements are of some standard composition? Hope you don't mind my asking.

I pick up bits an pieces when I ask questions that help me get over bits of ignorance I have collected as my interests change.

This link might help:

Newton's Universal Equation for Gravity

Those calculations look like a GA to me.

Hi Hero, you seem to have forgotten that the 2500 is km and not meters.

You are quite right that relativistic effects are negligible here.

Jorrie

Ha! You are right. That explains why I was off a few decimal places!!! :)

Exponent should be 10^23 after rerunning the numbers.

Shouldn't that be 2.634 X 10^17 given that you've done the calc. for a planet which is 1000 times too small.

Actually, no. Here is why...

First, the units are in meters and kilograms. I mixed meters with kilometers for the radius. The problem with that was in the Universal Gravity Equation calculation:

a = G*M/r^2

We know a, G, and r, but need to solve for M. So we rearrange the equation to:

M = (a * r^2) / G

Since the number I used for radius was off by 1,000 (2,500 versus 2,500,000), that means r^2 should have been a much larger number than the value I initially calculated (6.25 E12 instead of 6.25 E6). So, you can see the error is in the numerator and the magnitude of the squared radius was too small.

That means the end result with divisor (G) would yield a smaller number than it should have been by four decimal places (E19 vs. E23).

Notwithstanding that, Randall's right. The calculation in #2

2.8125 m/s/s = 6.6743 X 10^-11 * M / 2500^2 ==> M = 2.634 X 10^19 kg is wrong.

It should be M = 2.634 X 10^17 kg i.e. down by a factor 10⁶.

Cheers.........Codey

Now I understand. Thanks, nothing like peer review. ;-)

I believe that you skipped some powers of ten. The radius of the planet is 2500 klm which means 2.5x10^6 m.

I agree with your logic, I calculated this the same way. I believe you should have converted R to meters. I used an average value of R=2,502,000 m. I got a mass of 2.638 x 10^23 kg.

One important piece of information that has not been provided is the presence and properties of an atmosphere. An atmosphere will create a potentially significant drag force, which will slow the projectile (ejecta) and result in a lower ultimate height than if it were projected in a vacuum. This reduced height will result in an artificially high calculation of the acceleration at the surface (a = v² / 2h). For the given height of 4000 m (assuming in atmosphere), the calculated surface acceleration is 2.8125 m/s², but if for example the height in vacuum increased to 5000 m, then the true surface acceleration is only 2.25 m/s².

Calculation of the drag force and the resulting decrease in ultimate height is a complex procedure and depends on variables such as the density & viscosity of the atmosphere, shape of the ejecta and initial velocity. This shows the concept of reduced height and range for a projectile in atmosphere.

So, without any information given about the properties of the atmosphere (or lack thereof) I would say the answer is "No, it is not possible to determine the mass of the planet". However, if it is assumed that there is is no atmosphere, then I think Jorrie's & Hero's answer of 2.635x10²³ kg is correct. If an atmosphere is present, then this value would represent the maximum mass of the planet, since the true surface acceleration would be lower than the calculated acceleration and per Newton's Theory of Universal Gravitation (after some rearrangment of terms), M = (a · R²) / G, so the true mass will be lower than the calculated mass.

Like most of these chalenge questions it is necessary to make some assumptions. In this case it is not posible to use the information given to provide the planets mass with a high degree of accuracy unless some assumptions are made, ie, negigible atmosphere, planet has negligible rotation at location of volcano (earth has approx 0.5% less apparent gravity at equator than at poles due to centrifugal effects from its rotation), negligible height of volcano from which lava ejected, and perfectly spherical planet. Given all of these the answer provided by Anon Hero is correct.

misread it

I agree that several assumptions must be made, for which many are negligible. However, assuming negligible atmosphere is a pretty big leap of faith. If I am visiting this new planet and intend to land on the surface, I don't want my entry trajectory determined by the acceleration or planetary mass calculated based on negligible atmosphere when in fact there is heavy atmosphere (or vice versa).

Yes, with all the usual assumptions about no air resistance (if there is an atmosphere) etc.

I make it about 2.6*10²³ kg.

Cheers..........Codey

Not unless you make a bunch of assumptions not supported by the information supplied.

You did not ask that we try to determine the mass, rather whether it is possible to do so with the information given.

I believe the most correct answer here is... a qualified "yes".

One could probably arrive at a close approximation of the mass.

Actually the question was "is it possible to determine the mass with the info provided" The answer is NO. And the calculations for a zero atmosphere planet would give a minimal mass not a maximal mass.

No, it would be neither minimal nor maximal.

The presence of a thin cold atmosphere would cause drag that would slow the volcanic ash down, and not provide sufficient buoyancy to compensate. Under those conditions the mass of the planet would be lower than these calculations suggest.

On the other hand, planetary rotation means that additional (downwards) centripetal acceleration would be required, which would require the planet to have greater mass.

Just a couple of relatively straightforward examples.

No, it is the maximum mass. Atmospheric drag will cause the ejecta to achieve a lower overall height. In order to achieve the same height as for the zero-atmosphere condition with the same initial velocity, the surface gravity needs to be less and thus less planetary mass.

Anonymous Hero kindly referred me to further reading.

I posted my understanding of the question thinking all along this was a "Good enough" scenario.

It made sense to me that to simplify the equation calculations made from Center of the Mass would come as close as you might expect to get sans actual on the day at the place measurements and weights.

I thought that the utility of the equations was it did appear to enable "Good Enough" explorations, as was accomplished by Satellite and Robotic missions prior to actual attempts to land on someplace like the Moon or Mars.

Further I felt that the calculations dependent on centering would naturally include the over all mass, which would naturally include whatever atmospheric limits the "Body" or mass has.

It is unlikely that I will ever get to a competitive level with the math.

And you can be absolutely certain I have no expectation of ever getting a GA from any post I might make on this subject.

I suppose you are correct, there probably will never be sufficient info to determine the actual mass of the planet. All we can do is quibble over how closely we can estimate the mass. The more data available, the closer the estimate.

In the Newton gravitational equation as applied to this problem, r should be 2504, not 2500.

The equation F = GM_pm_a/r² (where Mp is the mass of the planet and m_a is the mass of the lava) tells us that r is distance between the center of the planet and the highest point attained by the lava.

In a typical volcanic eruption, all the lava wouldn't rise to 4 km. Some of it would fall back well before reaching this height. Here, I'm assuming all the lava escapes from the volcano in one shot - as if it were a cannonball. This cannonball reaches a height of 4 km.

Therefore, r becomes the distance from the center of the planet to the center of the cannonball, or 2504 km.

It would make a very minor difference to the final answer, but never let it be said we didn't think of everything.

Neither the lower figure nor the higher one is correct. You need to integrate the potential gravitational energy over height. That is ∫constants/R².dR. For such a small range, taking the midrange value for the gravitational acceleration (i.e the acceleration at 2502-km) would be extremely accurate, albeit still not theoretically exact.

Actually, I think it is the integral of 2500 to 2504 km since the object transverses through those altitudes. Silly as it may be, the initial altitude is actually the height of the volcano, which is probably above the average radius of 2500 km. But...

In the end, 2504 vs 2500 represents less than .2% error, which is lost in the noise if you ask me.

And... For everyone talking about aerodynamic drag, it is also probably lost in the noise (or wind if you prefer). Here is my reasoning:

Compare our fictitious planet (let's call it mystery planet X) with, say, Mars. Mars has a radius of 3400 km and a mass of 6.6 X 10E23 kg. That is more than 2 times planet X's mass and we all know that Mars doesn't have much of an atmosphere (sort of like some local bars around here, too).

Planet X has a radius about that of Mercury, which is just about 2400 km. Hmm, even less of an atmosphere there, but we can blame the solar wind for that. Our own Moon has a radius of about 1740 km, but its mass is a little less than 1/3 planet X.

Regardless, a little common sense tells us that with a radius that small it is unlikely to have much of an atmosphere and therefore the aerodynamic drag component is also probably low enough to consider it not important in the overall scheme of things. That was my reasoning in the beginning.

Titan (a moon of Saturn) has a diameter of roughly 2500 km, yet it has an atmosphere about 1.5 times higher pressure than that here on Earth. NASA also discovered fairly recently a significant atmosphere on another Saturn moon, Enceladus, which is only 500 km diameter.

From the NASA JPL:

"Enceladus is a relatively small moon. The amount of gravity it exerts is not enough to hold an atmosphere very long. Therefore, at Enceladus, a strong continuous source is required to maintain the atmosphere.

The need for such a strong source leads scientists to consider eruptions, such as volcanoes and geysers."

Thus, if this new planet has significant volcanic activity, it could indeed have an atmosphere that needs to be considered.

Good research! Interesting to note the atmospheric density. I would not have expected that!

the mass is proportional r*r*r, so the double radius will need the 8times mass!

compare the planet X with the planet pluto (he shall have an atmosphere)!

If the density is the same, double the radius would provide 8x the mass. But it would also have twice the gravity at the surface. If we were to doubly the planet's radius, but keep all the other measures for the lava the same (including any that are not stated), the planet would have to be about 4x the mass.

If the planet is not from gas the matter should incompressible (the gravitational force has no effect to the density).

Does this mean anything?

at normal rates (pressures and temperatures) this is for any hard material and most liquids!

Yet another apparently obvious* statement without explanation of relevancy.

It went without saying that your calculation that doubling the radius caused x8_mass/x2_gravity was based on an approximation of constant density; but the comment which prompted your apparent non-sequitur said that if we wanted to maintain the same surface gravity the mass could only be x4. That would merely require the planet to be made from lighter materials - no great problem there.

Regarding your comment on gaseous planets - the question's reliance on the surface of the planet (not to mention the existence of volcanoes) implies that the material immediately below is either solid or liquid.

BTW, I think we could theoretically achieve the required density in an all-gas planet of this size - but it would need to be constituted of dense gas (such as Xeon or SF₆), and a very long way from the nearest star, so that the temperature at the "top" of the atmosphere would be close to 2-K. (But don't ask how such a planet might come to exist in the first place).

*But actually incorrect: solids are far from incompessible under normal (planetary) conditions; iron is a relatively incompressible solid, but even the earth's core is under sufficient pressure as to increase the s.g. of iron from just under 8 to somewhat over 12.

simple calculation:

radius is 2500km equals 2500000m, mass is 2.63E23kg, volume is 4*pi*r*r*r/3, density is nearly m/(4*r³) = 2.63E23/61E18m³ = 4.3 times the density of water!

This is a theoretical value and is an average of all materials in that mass, light and heavy materials, solids, liquids and gaseous!

The gas maybe in porose materials included or as an atmosphere exist!

The molecular spacing in a compressed gas need be no greater than in a solid of similar molecular dimensions. All you need is enough pressure.

this high enough pressure on molecular gases will result in a gas density nearly the same value of solid materials (normal air has a density of 1.3kg/m³ - iron has a density of a bit less 8000 kg/m³); the difference between a gas like air and a solid material like steel is a faktor of 8000, the factor of iron on the surface and in the core is 1.5 or 2!

That's a little difference!

The density of the sun kis much lower thean the density of water!

the density of 12 for iron in the core of the earth is lighter then uranium on the surface of the earth!

There are two variables which affect the calculation. Atomspheric density has already been noted. Variable #2 is the intial velocity of the ejector material. Vulcanism is not an accurate tool for calculating planetary mass.

Assuming this plant is orbiting a star - there are better astronomical calculations that will provide the mass.

Carolynne

Apart from the more direct mass measurement that I noted in reply #40, the easiest for the spacefarers would probably have been to orbit the planet a few times (in free-fall mode) before landing. Assuming that they have the precision radar altimeter required for the landing, they could determine the semi-major axis (a) of their orbit and their orbital period (T) with good accuracy. The mass of the planet is then simply given by Kepler's third law as:

M = 4Π²a³ / GT²,

assuming a perfectly spherical planet. I suppose they could have used optical means to determine any oblateness of the planet and compensate for that by means of a more complex calculation.

-J

Indeed, the challenge method would only be appropriate for an initial 'dirty' estimate at a time when you do not have direct access to the planet.
It's sole advantage is that (given that a Volcano is currently active) results could be generated optically and quickly. Anyone visiting the planet would already have data available that allowed a much more accurate estimate (otherwise they could not have landed safely).

The initial speed of the lava = 150m/s

The Final speed of the lava = 0m/s

The total displacement = 4km = 4000m

The radius is = 2500km = 2500000m

First we have to find "g" which is the planets accelleration so since the lava is spewing up to a height of 4km (assuming the planet is a perfect sphere) with an initial speed of 150m/s with a final velocity of 0m/s. Formula of motion:

V²=U²+2as

0² = 150² + (2 xa x4000)

0 = 22500 + 8000a

-22500 = 8000a

-2.8125m/s² = a = g the negative sign stands for the fact that the accelleration due to gravity is acting torwards the planet and that the lava is decellerating. we are also assuming that the accelleration is constant for this distance of 4km.

g = GM/r² where G is the Universal Gravitational constant with a value of 6,67 x10^-11 M stands for the value of the planet's mass and r stands for the radius of the planet. since we want M we make it the subject of the formula:

M = gr²/G = 2,8125(2500000)²/(6,67 x10^-11) =

2,6354 x10^23kg is the estimated mass of the planet which is smaller than planet earth.

To answer the question yes this information can be used to find the mass of the planet.

I'm new in this discussion, but as a Computer Scientist Eng. I just conducted a non constant gravity simulation of the problem, here comes the result:

The exact mass of the planet should be:

Kg.

Bigger figures than reportet previously.

If interested in the algorithm details, please find It here (using Visual Basic):

g = 6.67428E-11
v0 = 150
hmax = 4000
radi = 2500000
Mplanet = 2,638915e+23
a = 0#
h = 0#
v = v0
t = 0#
dt = 0.01
While (v > 0)
t = t + dt
h = h + v * dt + 0.5 * a * dt * dt
d = radi + h
a = g * Mplanet / (d * d)
v = v - a * dt
Wend

Once v equals 0 (lava stops), h is compared with 4000 m, if h is higher then a bigger mass planet is used, otherwise an smaller mass planet is used instead, until h converges to 4000 m.

Hope it helps,

You make a good point. Given the precision of the information provided, the error associated with just the simple mass calculation is about ±15%. I have used the root-sum-square (RSS) method for determining this error, although if I use the max and/or min data values for each calc I get a maximum error of about 27%.

Since the data values are round, I have assumed that all zeros are not sig figs.

v = 150 ± 5 m/s

h = 4 ± 0.5 km = 4000 ± 500 m

R = 2500 ± 50 km = 2500000 ± 50000 m

G = 6.67428 x 10^-11 m³ / kg · s²

a = v² / 2h

= 2.8125 m/s²

M = (a · R²) / G

= 2.634 x 10²³ kg

RSS Method:

For w = f (x,y,z),

(Δw)² = (∂w/∂x · Δx)² + (∂w/∂y · Δy)² + (∂w/∂z · Δz)²

Thus,

Δa = ((v/h · Δv)² + (-v²/2h² · Δh)²)^½

= 0.398 m/s² (±14.2%)

and

ΔM = ((R²/G · Δa)² + (2aR/G · ΔR)²)^½

= 3.88 x 10²² kg (±14.7%)

As regards the limits of precision, I agree your principle - though perhaps it's reasonable to assume that the height would have been stated to 2 sig figs. like the other parameters.

I assume that you are aiming for an rms figure, in which case a reasonable assumption would be that rounding errors are uniformly distributed. I believe that this means the result should be a factor of 1/√3 smaller.

Other than that, my crude method based on your (probably worst-case) assumptions would give the same results as yours:
σ(V)/V = 5/150/√3 and dM/dV/(M/V)=2
σ(h)/h = 0.5/4/√3 and dM/dh/(M/h)=1
σ(R)/R = 50/2500/√3 and dM/dR/(M/R)=2
which gives an RMS error for the mass of 8.5%

(If it's peak error you are after, you need to take asymmetries into account - easiest done by calculating the extrema - and with only three linearly distributed parameters the probability of getting near the extrema is not ridiculously low)

I'm not sure that RMS is appropriate. I was simply calculating the propogated error of the precision uncertainty on the mass calc.

.

Actually, maybe peak error is more appropriate for the point I was trying to make. Using extreme values of v=145 m/s, h=4.5 km and R=2450 km, the calculated mass is 2.10 x 10²³ kg (-20%), while using values of v=155 m/s, h=3.5 km and R=2550 km results in 3.34 x 10²³ kg (+27%).

The RSS method does use the sum of squares, so doesn't this account for the probability of all parameters deviating to extreme values at the same time?

RSS is only meaningful for the summation of effects that are already measured as standard deviation (or RMS variation) - that is intrinsic to its structure (square, then sum, then square-root).
As regards probability of a an error of a given size: RMS/standard_deviation as a single figure is only meaningful if the distribution is normal (or Gaussian). The sum of three uniformly distributed variables with one variable significantly larger than the others will be more similar to a uniform distribution than to a Gaussian one.
So: RSS is not meaningful in this case, although it could be if the calculation of this mass was merely part of a larger calculation which involved a number of variables with greater total contribution to the final outcome (and you would need to start with RMS values). (The aggregate mass of a multi-planetary system, for example?) Though of course you would not choose this method of measurement if you had an alternative available.

I can't follow either your error analysis or that of FYZ's reply

We have M = (V².R²)/(2.h)

Take logs ln(M) = 2.ln(V)+2.ln(R)-ln(2)-ln(h)

differentiate dM/M = 2.dV/V + 2.dR/R -dh/h

for small departures this becomes ΔM/M = 2.ΔV/V + 2.dR/R -Δh/h

where each term represents the "extreme error" caused by the "variable". In this case

ΔM/M =(%M) = 2*5/150 + 2*50/2500 + .5/4 = 6.67% + 4.00% +12.5% = 23%

using a square weighting gives (.44% + .16% + 1.5%)^½ = 15%

I was just trying to show that using extreme values (max velocity, max radius, min height) that the calculated value could be up to 27% higher than the expected value. The error that I calculated in #47 using RSS is the same as you have presented here.

where have you been to school?

you are a prophet?

The planetary mass is searched for!

Very impressive solution, however, we have lost sight of the question asked. The correct answer to the quesiton is: Yes!

Hi Guest, with your unqualified "Yes", you would not have scored more than 30% of the marks for the question in an exam, depending on the examiner, I guess.

The information given only allow for a very rough-cut estimation of the real mass of the planet, as many respondents have described above.

-J

The answer is a clear No! You must know something about the atmosphere. This is not a ballistic problem; there is considerable convective action in a normal volcanic eruption of this sort (see Pliny for details). The lava ejected high above the cone is no longer great globs of molten rock but quite pulverized. Heating of the atmosphere along the sides produces a chimney effect that increases the height.

GA - this certainly would add complexity to the solution (if atmosphere is present) and make it only partially dependent on the mass of the planet.

Simply, NO.

Variables are missing. Such as atmospheric density and diameter, composition and density of lava erupted (viscosity and element/compound chemistry), gravity strength of planet (yeah okay, pushing it with this one) and more importantly...the force of expulsion (magmatic pressure versus diameter of volcanic vent expulsing the lava).

...not all lavas are alike

...not all volcanoes are alike

...not all planets are alike

"Variables are missing. Such as atmospheric density and diameter ..."

The author has the right to formulate the problem any way he/she sees fit and to include or omit details as he/she sees fit. The problem's formulation is completely at the whim of the author, as it should be. These variables are 'missing' because the author considers them irrelevant, perhaps?

If it were my problem I would have used something else. A 2500-km blob of Jell-O, maybe. That would make everyone happy because, as everybody knows, big blobs of Jell-O don't come with atmospheres (but you can order them with whipped cream!)

"... composition and density of lava erupted (viscosity and....."

Once the lava (or anything else) is ejected it is in free-fall and its physical properties are of no consequence. The point is that the stuff has an initial velocity of 150 m/s. That's all that matters. The author could just as easily have ejected something else at 150 m/s. Tribbles, for example. Or Easter eggs. It doesn't matter.

"...not all planets are alike"

Exactly. Some of 'em don't even come with atmospheres, volcanoes or anything else. Not even Tribbles.

Care for some whipped cream?

lol....those points should have been clarified before formulating a response for a statement

Hey - had your joint today?