Physics question

unclebutcher SG

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

^tamago^ SG

up…. first qn seems like 1/9w.

eagle SG

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

unclebutcher SG

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.

eagle SG

Density is mass over volume

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...

unclebutcher SG

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

eagle SG

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

teraexa SG

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 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.

unclebutcher SG

Originally posted by teraexa:

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]

thank you very the much.....your physics damn pro sia

eagle SG

Evil or Very Mad

Never thank me for my last one

And erm... This is simple gravitation btw... -.-

unclebutcher SG

im just damn confused over the helluva lot of formulae in gravitation....and physics...rather do chem Wink

eagle SG

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.

unclebutcher SG

-gmm/r^2
gm/r
t=(2pi)/omega
gmm/r
etcetcetcetcetc

Mad Mad Mad Mad Mad

eagle SG

careful

g and G are different here.
So is M and m

teraexa SG

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.

teraexa SG

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.

unclebutcher SG

Originally posted by teraexa:

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.

my kind of subject.

teraexa SG

Wait till you get to physical chem. A pure mug will net you a B at most.

unclebutcher SG

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....

teraexa SG

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.