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Help calculating mass and escape velocity
Monkey
post Jan 21 2008, 07:47 PM
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If you have a kuiper belt object with say a diameter of 48km and assume a density of 1.5 gm/cm^3, how do you find the mass and escape velocity?

Sorry if this sounds elementary, I am just trying to learn. I think that the surface area would be 7,238km^2 (4pi(r^2)) and that the volume would be 57,905km^3 ((4/3)*pi*r^3). I seem to start getting into trouble solving for the mass (given that density = mass/volume.) Would the mass only be 1.5 * 57,905,000m^3 = 86,857,500m^3 = 86,857.5 kg? What would the escape velocity be?
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ugordan
post Jan 21 2008, 07:51 PM
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Watch out for the unit conversions. A common mistake is assuming 1 km^3 is 1000 m^3. It is equal to 1000x1000x1000 m^3 which translates into a lot of mass. The escape velocity is then SQRT(2*G*M/r), see Wikipedia for details.


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Greg Hullender
post Jan 21 2008, 08:03 PM
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I guess we're assuming this KBO is a sphere. :-)

The surface and volume calculations are correct; you get into trouble trying to use your density (which is g/cm^3) with your volume (which is km^3). Also, although there are 1,000 m in a km, there are 1,000,000 m^2 in a km^2 and 1,000,000,000 m^3 in a km^3. Since your density is in cm^3, you'll need to go from km^3 to cm^3 before you can multiply it by the density, and your result will be grams.

Escape velocity is sqrt(2MG/r) where M is the mass of the KBO, G is the gravitational constant, and r is the radius. You cannot derive this formula without Calculus, but if you do know Calculus, it's not hard. Again, you'll need to get the gravitational constant into the same sort of units as you're using for M and r.

Hope that helps.

So is this for a class, or just your own curiosity? :-)

--Greg
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JRehling
post Jan 21 2008, 08:03 PM
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I always liked the chestnut that escape velocity is proportional to density times diameter. If you know the EV of one world, and you can express the second world's density and diameter in terms of the first one, you can work the answer that way. For example, the Moon has about 3/5 the density of the Earth and about 1/4 the diameter, so it has roughly 3/20 = 0.15 the escape velocity. If you're trying to save time at the expense of significant digits... wink.gif
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Greg Hullender
post Jan 21 2008, 08:31 PM
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By the way, I recommend against a 55-mile-per-hour speed limit on this KBO.

--Greg :-)
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Monkey
post Jan 21 2008, 09:50 PM
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QUOTE (Greg Hullender @ Jan 21 2008, 03:03 PM) *
I guess we're assuming this KBO is a sphere. :-)

The surface and volume calculations are correct; you get into trouble trying to use your density (which is g/cm^3) with your volume (which is km^3). Also, although there are 1,000 m in a km, there are 1,000,000 m^2 in a km^2 and 1,000,000,000 m^3 in a km^3. Since your density is in cm^3, you'll need to go from km^3 to cm^3 before you can multiply it by the density, and your result will be grams.

Escape velocity is sqrt(2MG/r) where M is the mass of the KBO, G is the gravitational constant, and r is the radius. You cannot derive this formula without Calculus, but if you do know Calculus, it's not hard. Again, you'll need to get the gravitational constant into the same sort of units as you're using for M and r.

Hope that helps.

So is this for a class, or just your own curiosity? :-)

--Greg


Hi Greg;

I am a freshman in high school, so no calculus yet. So if I am doing the mass calculation correctly:

Given a basically spherical KBO that is 48km in diameter with an assumed density of 1.5 gm/cm^3 and thus volume = 57,905km^3; Mass = 1.5 * 57,905km^3 = 86,857.5km^3. (km^3 conversion: 1000^3 = 10^9m; Further conversion to cm = another 10^2) so the mass = 8.68575 x 10^15kg? (9+2+4) or am I still off by a factor of 10-1000?

Help!
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alan
post Jan 21 2008, 10:17 PM
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QUOTE (Monkey @ Jan 21 2008, 03:50 PM) *
Hi Greg;

I am a freshman in high school, so no calculus yet. So if I am doing the mass calculation correctly:

Given a basically spherical KBO that is 48km in diameter with an assumed density of 1.5 gm/cm^3 and thus volume = 57,905km^3; Mass = 1.5 * 57,905km^3 = 86,857.5km^3. (km^3 conversion: 1000^3 = 10^9m; Further conversion to cm = another 10^2) so the mass = 8.68575 x 10^15kg? (9+2+4) or am I still off by a factor of 10-1000?

Help!

86,857.5km^3 should be in g*km^3/cm^3

convert g to kg ( /1000); convert cm^-3 to m^-3 (/100^-3); km^3 to m^3 (*1000^3)

exponent = 4-3+(3*2)+(3*3) = 16
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Greg Hullender
post Jan 21 2008, 10:44 PM
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Monkey, there's a special technique called "Dimensional Analysis" that lets us keep track of all these different units. It's a lot easier to show on a whiteboard, though; it's hard to draw a line through things we want to "cancel out." Here's a shot at it, though.

First, obviously you know that mass is density times volume. But this time let's take your original expression and see what we can do with it.

We start with what you started with:

(1.5 g/cm^3)(57,905 km^3)

As you can see, we have volume(km^3) in the numerator and we also have volume(cm^3) in the denominator, so they ought to cancel out and just leave us with mass (g). However, you're not allowed to cancel km^3 against cm^3; you can only cancel km against km and cm against cm.

But we DO know that (100 cm/1 m) = 1 and (1000 m/1 km) = 1. Also, (1 kg / 1000 g) = 1. Obviously we can multiply ANYTHING by 1 and get the same answer. So look at this equation:

(1.5 g/cm^3)(57,905 km^3)(100 cm/1 m)^3 * (1000 m/1 km)^3 * (1 kg / 1000 g) = (1.5)(57, 905)(100^3)(1000^3)/1000 kg.

See how I got all the g, m, km, and cm terms to cancel each other out? Nothing is left but the kg term. That works out to 8.6857E+16 kg.

Dimensional analysis is very easy once you get the hang of it, but it's very, very important in science and engineering.

--Greg
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Monkey
post Jan 22 2008, 12:26 AM
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QUOTE (Greg Hullender @ Jan 21 2008, 05:44 PM) *
But we DO know that (100 cm/1 m) = 1 and (1000 m/1 km) = 1. Also, (1 kg / 1000 g) = 1. Obviously we can multiply ANYTHING by 1 and get the same answer. So look at this equation:

(1.5 g/cm^3)(57,905 km^3)(100 cm/1 m)^3 * (1000 m/1 km)^3 * (1 kg / 1000 g) = (1.5)(57, 905)(100^3)(1000^3)/1000 kg.

See how I got all the g, m, km, and cm terms to cancel each other out? Nothing is left but the kg term. That works out to 8.6857E+16 kg.

Dimensional analysis is very easy once you get the hang of it, but it's very, very important in science and engineering.

--Greg


Hi Greg/Guys;

This is starting to make sense.

4 Start + 100^3 "6" + 1000^3 "9" + denominator "-3" = (4+6+9-3) = 16

Diameter = 48km
Density = 1.5 g/cm^3
Surface Area = 7,238km^2
Volume = 57,905km^3
Mass = 8.68575 x 10^16 kg


Now for the escape velocity sqrt(2MG/r), how do I convert the gravitational constant (G) to the units that I want to use for the numbers I have? Wikipedia shows 6.67428x10^-11 m^3 kg^-1 s^-2. The radius is 24,000m and the mass is 8.6x10^16 kg.

So is the escape velocity 21.87 m/s = 0.02187 km/s = 49MPH ?? (0.02187*3600/1.6)
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Greg Hullender
post Jan 22 2008, 12:46 AM
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Yes, 49 mph is what I got too. Like I said, 55 mph speed limit is way too high for this KBO. :-)

On the other hand, it means astronauts wouldn't have to worry about accidentally jumping off.

By the way, let me suggest that you do the experiment of converting the 21.87 m/s into miles/hr using only the (exact) metric to English conversion (1 in/2.54 cm) = 1. I see you got it right in a single step, but I think it'd be worth the practice to do it out. I take 5 or 6 steps to do it.

--Greg
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Monkey
post Jan 22 2008, 02:51 AM
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QUOTE (Greg Hullender @ Jan 21 2008, 07:46 PM) *
Yes, 49 mph is what I got too. Like I said, 55 mph speed limit is way too high for this KBO. :-)

On the other hand, it means astronauts wouldn't have to worry about accidentally jumping off.

By the way, let me suggest that you do the experiment of converting the 21.87 m/s into miles/hr using only the (exact) metric to English conversion (1 in/2.54 cm) = 1. I see you got it right in a single step, but I think it'd be worth the practice to do it out. I take 5 or 6 steps to do it.

--Greg


Greg thanks again for the help. The only other question I can think of is, how would I find the surface gravity in m/s of this 48km diameter KBO?
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Greg Hullender
post Jan 22 2008, 04:38 PM
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Sure, but first let me make another point about units. Gravity is m/s^2 -- not m/s. VERY important distinction.

Next, have a look at the formula for gravity at the top of this page:

http://en.wikipedia.org/wiki/Gravitational_constant

See if you can work it out from that.

--Greg
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qraal
post Jun 6 2008, 09:17 AM
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Hi Guys

"Monkey" has probably moved on well passed this question, but JRehling made an interesting observation. Since mass = density*volume, and volume is 4/3*pi*radius^3 (for a sphere), and surface gravity is G*mass/radius^2, thus gravity is 4/3*pi*G*density*radius. Escape velocity squared is 2*G*mass/radius, thus it is 2*gee*radius, which is all 8/3*pi*G*density*radius^2. Or 2/3*pi*G*density*diameter^2.

Thus escape velocity is directly proportional to diameter, but directly proportional only to the square-root of the density. Thus the Moon's escape velocity is roughly 4/5*1/4 ~ 0.2 of Earth. Jupiter's is roughly 11*1/2 = 5.5 Earth's, and so on. This doesn't tell us very much about the planets, but can be kind of fun to play with if we're teaching grade-school kids.


QUOTE (JRehling @ Jan 22 2008, 08:03 AM) *
I always liked the chestnut that escape velocity is proportional to density times diameter. If you know the EV of one world, and you can express the second world's density and diameter in terms of the first one, you can work the answer that way. For example, the Moon has about 3/5 the density of the Earth and about 1/4 the diameter, so it has roughly 3/20 = 0.15 the escape velocity. If you're trying to save time at the expense of significant digits... wink.gif

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