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View Full Version : Mass vs Gravitational Force... I know I am an Idiot.



Suds101
2008-Sep-26, 08:06 AM
For some reason, general Freshman year physics is escaping my mind and maybe its because I am up way to late studying for my ASTR. 100 test tomorrow, I can't rap my brain around this question. The question states:

Suppose two asteroids are located at the same distance from the Sun. One asteroid has twice the mass of the other. According to Newton's law of gravitation (and ignoring all forces except that from the Sun),

a. the more massive asteroid feels half the force that the other does.
b. the more massive asteroid feels twice the force that the other does.
c. neither feels any force because they are weightless in space.
d. both asteroids feel the same force because gravity acts equally on all objects.

Anyone, please? I know its either B or D but I cant feel confident on either answer.

Sock puppet
2008-Sep-26, 09:16 AM
Use Newton's equation for gravitational force: F=GMm/r^2
Anything more would just be telling you the answer.

trinitree88
2008-Sep-26, 10:54 AM
Use Newton's equation for gravitational force: F=GMm/r^2
Anything more would just be telling you the answer.

And for a rookie...some texts use m1...and m2 for the two interacting masses. Others use a particular subscript for the sun, with a circle with a dot, to indicate a solar mass. It really doesn't matter the order of multiplication,(commutative operation)...but i've seen enough rookies balk at plugging in the data, simply because of the notation. pete

for sun's mass see:http://www.enchantedlearning.com/subjects/astronomy/sun/sunsize.shtml

jlhredshift
2008-Sep-26, 12:01 PM
Remember the scene on the Moon when the Apollo astronauts dropped the hammer and the feather.

NEOWatcher
2008-Sep-26, 12:31 PM
I have a problem with the word "feel" and the point of view of the question.
Of course, Newton shows how the attraction can be different, and shows that the scale of the mass differences matters (thus the hammer and feather against the moon)
But; if it's the asteroid doing the "feeling", then does it's own gravity matter?
And there are 2 answers I would throw out right away, but not the same 2 because of the "feel".

cjameshuff
2008-Sep-26, 02:15 PM
And remember that force and acceleration are not the same thing, especially when considering the hammer and feather example. A = F/m

01101001
2008-Sep-26, 03:20 PM
Remember the scene on the Moon when the Apollo astronauts dropped the hammer and the feather.

I must be older. I remembered Galileo in Pisa hollering, "Bombs away!"

mugaliens
2008-Sep-26, 05:10 PM
e. both all and none of the above (ie, one or the other, depending on what precise moment the question is asked).

Although with a one-post name like Suds101, I'd wager money on this being a collegiate way of stumping the dummy.

Almost neither feels any force, but not because they're weightless in space. Rather, it's because gravity fairly uniformly acts on the asteroid as a whole.

Some slight difference in force is felt, however, as the part nearer to the Sun experiences more gravitational attraction than the further side. This results in the only force felt by the asteroid is one of a very slight stretching parallel to imaginary lines drawn towards the center of the Sun.

So, really, e. None of the above.

It's probably a simple case of your instructor failing to be precise in the wording of their multiple choice answers.

Jeff Root
2008-Sep-26, 11:11 PM
Almost certainly, the word "feels" is an unintentional red herring.

I think that tidal forces from the Sun on an asteroid can safely be
considered negligible.

The question appears to be about force, not about what is felt.

The question appears to be about force, not about acceleration.

That leaves only one possible correct answer.

You will be in a better position to complain about the provided answers
if you answer "correctly".

-- Jeff, in Minneapolis

hhEb09'1
2008-Sep-27, 10:33 AM
I agree with Jeff and the analysis, that is the target answer, probably.

Especially since most asteroids are cold, unfeeling wretches.

mugaliens
2008-Sep-27, 02:42 PM
I think that tidal forces from the Sun on an asteroid can safely be
considered negligible.

In this case, and given your other comment that it's about force, rather than what's felt, the answer is obviously that the force is proportional to it's mass, which leaves but one answer of the four provided in the OP.

And that's about as much as any can say without blurting out the answer!

Celestial Mechanic
2008-Sep-28, 04:35 AM
I must be older. I remembered Galileo in Pisa hollering, "Bombs away!"
And his graduate students dropping the objects from the tower. :)

Jerry
2008-Sep-28, 04:42 PM
None of the above is the most technically correct answer...the more massive ball could 'feel' either more or less force, depending upon your definition of 'feel'. Both are accelerating towards the sun at almost identical rates, because the mass of both objects is so much less that the sun, and the mass of the objects can be ignored when calculating the accelerations...however, the more massive object pulls the sun towards it, more than the less massive object, therefore the actual acceleration of the larger object will be ~1x10^284 less than the less massive body.

Then, there are also the relativistic effects to consider, and here again, the motion of more massive object will be retarded even further.

But if you define 'feel' as which objects acceleration requires the most force, you get a different answer; and if 'feel' means which object will detect acceleration; you probably get the answer the teacher was looking for. (I usually answered this question 'wrong', just on principle).

hhEb09'1
2008-Sep-28, 04:48 PM
None of the above is the most technically correct answer...the more massive ball could 'feel' either more or less force, depending upon your definition of 'feel'. Both are accelerating towards the sun at almost identical rates, because the mass of both objects is so much less that the sun, and the mass of the objects can be ignored when calculating the accelerations...however, the more massive object pulls the sun towards it, more than the less massive object, therefore the actual acceleration of the larger object will be ~1x10^284 less than the less massive body.I'm not sure I follow that. Could you expand on that? Why the larger body has less acceleration? In what sense?
(I usually answered this question 'wrong', just on principle).I've only heard that excuse twice, in thirty years of teaching. :)