1) Is there any meaning of "Weight of the Earth"? I think it has, but it will be different for different planets and stars since they have different masses, so they'll be attracting earth towards themselves with different magnitude of force because,
g(acceleration due to gravity) = √GM/R2, where M & R are the mass and the radius of the body that is attracting earth towards itself
If I am wrong then please correct me.
2)At noon, the sun and the earth pull the objects on the earth's surface in opposite direction. At midnight the sun and the earth pull these objects in the same direction. Is the weight of an object as measured by the spring balance on the earth's surface more at midnight as compared to its weight at noon?
3)An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration g?
4)A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on earth? If yes then where? If no, how can an astronaut make it land on the earth?
Well the nut will land or not on the earth depends on where it is located, it will surely land on the earth if its inside the satellite( yeah I agree that, for that satellite too has to land on the earth). It is also depended on where satellite is located, if it is located far away from the earth and the nut is out side it and where the gravitational force is only optimum may be then nut will fly off in the space.
Don't know how will astronaut will make it land on earth. :P:P So please help. :)
11they havn't mention anything about mass, and I am assuming that mass is also decreasing,
your answer is correct only when mass remains constant. But they have said that the shell is maintaining its shape, it doesn't implies that its mass is not reducing.
71For 7, Why does my heart tell that it should be b). Let me know your opinion cause you might have thought too.
And 8 is correct too I think!
11why can't option (a) be correct as well ??
Give the reason for your answer (waise answer HCV mein bhi diya hai, par I need the reason)
71why can't option (b) be correct as well ?? Wahi toh likha hai. But I didn't know the answer, I haven't reach upto there. :)
For the 8th one, Simply because 'R' decreases thereby the the Gravitational Potential Energy decreases.
11sorry I meant why option (a) be correct. I edited it.
for it I think,
Assuming the Earth is exactly spherical, we expect gravity to always point towards the
center of Earth. However, the centrifugal force is perpendicular to the axis of the Earth.
Except on the equator, therefore, it is not exactly opposed to gravity, but adds a small
horizontal vector component, pointing towards the equator. As a result, not only is effective
gravity weakened, but its direction is modified--instead of pointing to the center of the Earth,
is slants (ever so slightly) towards the equator.
Though not sure if it is correct explanation to it.
71Well, I would understand those more better when I read that. Actually that chapter is ...
11Potential is - GM/R if the R decreases, that value too will decrease which implies that potential
shall increase, because it is negative and you know the smaller the number is with negative
sign the more is its value. So your reason is wrong I guess.
71GM/R,
When R decreases GM/R increases.
Now (GM/R) is increasing , so putting a negative sign in front of it will make it decreasing
So Gravitational Potential Energy decreases!
118)When a uniform spherical shell shrinks gradually maintaining its shape, why does gravitational potential at the centre decreases ?
71You don't need to think that much!
shell is maintaining its shape, signifies it maintains it rigidity and volume. and How can you decrease MASS (Which is always conserved).
49why do people avoid being called a scholar :P
u r indeed a scholar Vivek ;)
11Well Subhomoy is a genius.
I called Scholar to you Vivek, janbhoojkar anjan kyun bante ho?
Yeah Subho he is a great scholar. :)
1Ref. to Q.2 in Khyati's original post the weight of a substance is mg where 'g'= GM/ r2 Out of which 'G' is universal const and 'M' is mass of earth. Then how can the weight of the body depend on the attraction between earth and moon or sun? Can u explain?
71Then how can the weight of the body depend on the attraction between earth and moon or sun?
Well, It doesn't. That is what Subhomay has said.
11Ref. to Q.2 in Khyati's original post the weight of a substance is mg where 'g'= GM/ r2 Out of which 'G' is universal const and 'M' is mass of earth. Then how can the weight of the body depend on the attraction between earth and moon or sun? Can u explain?
g = √GM/R2,( we can use this formula to find the acceleration due to gravity of any
celestial body, provided we know its mass M and radius R)
Sengupta Ma'am,
all the celestial bodies has there own acceleration due to gravity, so they attract the object
towards themselves with different magnitude of the force.
I nowhere mentioned in my question about the dependence of the weight of the body
on the attraction between the earth, the moon or the sun.
If you still have any Ma'am please tell and always be there to ask question because it will
help us
:)
494) thought on this over-night....
the satellite around earth exists because it is present in the gravitational field of the earth...
and the satellite moves with a certain velocity which in its rotating frame of reference the centrifugal force balances the created gravitational force...
this, is true for each and every particle on the satellite...
even the nut...
so, i think no nut which opens up will go anywhere...it will just move with the satellite in its own position...
the astronaut can make the nut fall on earth just by giving the nut a small velocity towards the earth...
491)
why do u go till the planets?
weight of earth will differ man-to-man
for a man of mass 50 kg, weight of earth is 50g N for a man of weight 60kg weight of earth is 60g N (if weight is force with which the man attracts the earth toward's his centre of gravity...)
492)
well a wacky solution coming to my mind...
during noon, moon and sun attract the object in opposite directions with somewhat equal magnitude of forces,
during midnight also same is the case...
thus noon or midnight, the weight of the object will be the force exerted on the object by the earth ONLY!
dont know if i am correct or wrong! :P
493)
frame-of-reference! :P
the force exerted is the pseudo force :P
496) let the angular velocity of the earth be ω...
then, in the rotating frame of reference of earth, the net force experienced by a body of mass m is
mg-mω2R which is NOT zero!!
if this had to be zero then ω must have been √gR which is not the case!!
71If mw2r > mg then? He should fly out? Is it the case or not? Please let me know~
49exactly...u r right..
had mw2R been greater than mg then the object would surely have flown off...
but unfortunately/fortunately that value of w is also not present!! :D
otherwise imagine a lifeless portion of the earth in and around the equator [3]
115)Is it necessary for the plane of the orbit of a satellite to pass through the centre of the earth?
6)As earth rotates about its axis, a person living in his house at the equator goes in a circular obit of radius equal to the radius of the earth. Why does he/she not feel weightlessness as a satellite passenger feels?