Physics question
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On earth the weight of a particular object is w at the surface.assuming the earth radius to be r and mass to be m.what is the weight of the object at a point
3r from centre of earth?
On the surface of another planet radius 4r and mass 2r?
The question is related to gravitation -
up.... first qn seems like 1/9w.
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F = GMm/r^2 = wOriginally posted by unclebutcher:On earth the weight of a particular object is w at the surface.assuming the earth radius to be r and mass to be m.what is the weight of the object at a point
3r from centre of earth?
On the surface of another planet radius 4r and mass 2r?
The question is related to gravitation
as we can see, w is proportional to 1/r^2
hence, at a point 3r from earth, the weight will be w/9
another planet of mass 2r? I guess you mean 2m
F = GM(2m)/(4r)^2 = w/8 -
Thanks. Now i have got another question.Originally posted by eagle:F = GMm/r^2 = w
as we can see, w is proportional to 1/r^2
hence, at a point 3r from earth, the weight will be w/9
another planet of mass 2r? I guess you mean 2m
F = GM(2m)/(4r)^2 = w/8
Assuming that the earth is a uniform sphere.calculate th density of the earth im terms of G.the gravitational constant and r the radius of the earth. -
Density is mass over volumeOriginally posted by unclebutcher:Thanks. Now i have got another question.
Assuming that the earth is a uniform sphere.calculate th density of the earth im terms of G.the gravitational constant and r the radius of the earth.
volume = 4/3 pi r^3 = V
Now we know g = GM/r^2, M = mass of earth, and that g is a constant
hence M = g r^2 / G
hence,
Density = M/V = (3/4)(g / [pi r G])
P.S. not sure if g can be included like this... -
Thanks.Paiseh got another question. A star has radius 10km and mass 2.5 times ten power 29. A meteor is drawn into its gravitational field. Calculate the speed with which it will strike the surface of the star. Neglect initial speed of meteor. Thanks once again. Answer is 5.8 times ten power 7 ms-2
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ok
We must use the GPE formula P = -GMm/r
G = 6.673 * 10^-11
M = mass of star
m = mass of meteor
r = radius = 10000m
The change in GPE = gain in KE of the meteor
Hence
change in GPE = 0 - (-GMm/r)
gain in KE = 1/2 m v^2 - 0
Hence, GMm/r = 1/2 m v^2
Put all the values in
we get v^2 = 3.3365 * 10^15
Square-rooting gives u the answer -
I am confused by your phrasing here.Originally posted by unclebutcher:Thanks.Paiseh got another question. A star has radius 10km and mass 2.5 times ten power 29. A meteor is drawn into its gravitational field. Calculate the speed with which it will strike the surface of the star. Neglect initial speed of meteor. Thanks once again. Answer is 5.8 times ten power 7 ms-2
You mentioned speed but the answer u gave is ms-2, so I wonder which info is correct.
Anyway I shall just write the method of resolving the speed at which the meteor strikes the star.
Ok we know this is simply resolving the PE and KE of the system. Since the question told us to ignore the initial speed of the meteor, hence we can assume its initial KE to be 0. Also, at the start, we assume that the meteor is 'at infinity' from the star.
Recall your definition of gravitational potential as the amout of work done by an external agent in bringing an object to a particular point from infinity
So in this case, we can use this convenient expression: Loss in PE = Gain in KE. M is mass of star, m is mass of meteor.
At infinity, PE of system = 0 (by definition). PE at the end = -GMm/r where r is the radius of the star = 10 km.
Therefore 'gain' in PE = -GMm/r - 0 = -GMm/r. At here, we just take the magnitude of this value since we want the loss in PE.
Since loss in PE = gain in KE, we have this expression:
GMm/r = 1/2 m v^2
Cancelling m, we get v = sqrt(2GM/r) = sqrt(2 x 6.67E-11 x 2.5E29 / 10E3) = 5.77E7 m s-1. -
Originally posted by teraexa:
thank you very the much.....your physics damn pro sia
I am confused by your phrasing here.
You mentioned speed but the answer u gave is ms-2, so I wonder which info is correct.
Anyway I shall just write the method of resolving the speed at which the meteor strikes the star.
Ok we know this is simply resolving the PE and KE of the system. Since the question told us to ignore the initial speed of the meteor, hence we can assume its initial KE to be 0. Also, at the start, we assume that the meteor is 'at infinity' from the star.
Recall your definition of gravitational potential as the amout of work done by an external agent in bringing an object to a particular point [b]from infinity
So in this case, we can use this convenient expression: Loss in PE = Gain in KE. M is mass of star, m is mass of meteor.
At infinity, PE of system = 0 (by definition). PE at the end = -GMm/r where r is the radius of the star = 10 km.
Therefore 'gain' in PE = -GMm/r - 0 = -GMm/r. At here, we just take the magnitude of this value since we want the loss in PE.
Since loss in PE = gain in KE, we have this expression:
GMm/r = 1/2 m v^2
Cancelling m, we get v = sqrt(2GM/r) = sqrt(2 x 6.67E-11 x 2.5E29 / 10E3) = 5.77E7 m s-1.[/b] -
Originally posted by unclebutcher:thank you very the much.....your physics damn pro sia
Never thank me for my last one
And erm... This is simple gravitation btw... -.- -
im just damn confused over the helluva lot of formulae in gravitation....and physics...rather do chemOriginally posted by eagle:
Never thank me for my last one
And erm... This is simple gravitation btw... -.-
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play cheat method: Throw all available equations regarding that topic into the question. Find out which values you have for which equations, then use those particular equations.Originally posted by unclebutcher:im just damn confused over the helluva lot of formulae in gravitation....and physics...rather do chem -
-gmm/r^2Originally posted by eagle:play cheat method: Throw all available equations regarding that topic into the question. Find out which values you have for which equations, then use those particular equations.
gm/r
t=(2pi)/omega
gmm/r
etcetcetcetcetc
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carefulOriginally posted by unclebutcher:-gmm/r^2
gm/r
t=(2pi)/omega
gmm/r
etcetcetcetcetc
g and G are different here.
So is M and m -
Aiyo, let me give a brief rundown of the formulas of gravitation here:
When you want to calculate the gravitational force between 2 objects of mass m and M
F = -GMm/r^2 (negative cos gravity is attractive)
OK then there's the other one: g = -GM/r^2
g is basically the force per unit mass of m acting at that point.
From here you can see a nice relation F = -GMm/r^2 = -GM/r^2 x m = g * m.
Viola! F = mg. What a familiar equation!
Next, you need to know the one for energy:
PE between 2 bodies = -GMm/r (Note the r is singly powered)
PE per unit mass of m at a particular point = -GM/r (Same concept as force)
Change in PE = final PE - initial PE = -GMm/r-final - (-GMm/r-initial)
Lastly, you just need to equate the centripetal force with gravity.
Take the magnitude of both equations, you obtain:
GMm/r^2 = mv^2/r = mrw^2 = mr(2pi*f)^2 = mr(2pi/T)^2
After some simple manipulation, you can derive your Kepler's Third Law:
T^2 is proportional to R^3. -
Trust me, physics is way easier to score compared to chem. Chem has calculations, stuffs to pure mug +++ while physics with equations you can pretty much assure yourself of a B or C.Originally posted by unclebutcher:im just damn confused over the helluva lot of formulae in gravitation....and physics...rather do chem
It's not that I hate chem. In fact I used to love chem until I screwed up in the olympiad and my chem teacher changed. Physics is so much easier to score and self-study compared to chemistry. -
Originally posted by teraexa:
my kind of subject.
Trust me, physics is way easier to score compared to chem. Chem has calculations, stuffs to pure mug +++ while physics with equations you can pretty much assure yourself of a B or C.
It's not that I hate chem. In fact I used to love chem until I screwed up in the olympiad and my chem teacher changed. Physics is so much easier to score and self-study compared to chemistry. -
Wait till you get to physical chem. A pure mug will net you a B at most.Originally posted by unclebutcher:my kind of subject. -
i dunno what is your definition of mug...but for me it is doing every tutorial question given to me...every question asked i will answer...everything must understand or du lan...and finish tys....Originally posted by teraexa:Wait till you get to physical chem. A pure mug will net you a B at most. -
To me mugging is akin to stuffing stuffs into your brain without understanding it and regurgitating everything out in the exam hall.Originally posted by unclebutcher:i dunno what is your definition of mug...but for me it is doing every tutorial question given to me...every question asked i will answer...everything must understand or du lan...and finish tys....
Studying is to understand the concepts. Half the time students do not have a firm grasp of the concepts and hence they are unable to have a solid foundation to base their work on. -
ok...haha i should redefine...for me..i can understand chem easily so it is relatively easier for me...but not so with physics...Originally posted by teraexa:To me mugging is akin to stuffing stuffs into your brain without understanding it and regurgitating everything out in the exam hall.
Studying is to understand the concepts. Half the time students do not have a firm grasp of the concepts and hence they are unable to have a solid foundation to base their work on. -
Different inclinations I guess.
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heh heh... good...Originally posted by teraexa:To me mugging is akin to stuffing stuffs into your brain without understanding it and regurgitating everything out in the exam hall.
Studying is to understand the concepts. Half the time students do not have a firm grasp of the concepts and hence they are unable to have a solid foundation to base their work on.
you can manage the forums of questions while I do my FYP soon liao =D -
Buay sai.
Got A lvls this year.
That is a big killer. -
Haha... After your As before your BMTOriginally posted by teraexa:Buay sai.
Got A lvls this year.
That is a big killer.
Just nice my free time will be reduced to almost nil...