21 - Gravitation Questions Answers

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

Three particles A,B and C each of mass m are lying at the corners of an equilateral triangle of side L. If the particle A is released keeping the particles B and C fixed, the magnitude of instantaneous acceleration of A is

Asked By: SWATI KAPOOR
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Joshi sir comment

force by the other two on A = Gmm/L2 each but the angle between the two forces is 60 degree so resultant force = √3Gmm/L2

now calculate acceleration?

A large number of identical point masses m are placed along x-axis, at x= 0,1,2,4.........The magnitude of gravitational force on mass at origin (x=0), will be

Asked By: SWATI KAPOOR
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Joshi sir comment

The magnitude of gravitational force on mass at origin (x=0), will be

Gmm/12 + Gmm/22 + Gmm/42 + Gmm/8+ .........................

= [Gmm/12][1+(1/4)+(1/16)+....................]

= Gmm[1/(1-(1/4)]

solve?

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

A satellite of mass 200 kg revolves around a planet of mass 5 x 1030kg in a circular orbit of radius 6.6 x 106 m. Binding energy of the satellite.

Asked By: SWATI KAPOOR
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Joshi sir comment

Binding energy = GMm/2R

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

NAME  THE OUTERMOST LAYER OF SUN ????

Asked By: SARIKA SHARMA
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Joshi sir comment

i think its corona

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

A STONE WEIGHS 9N AT A PLACE ON THE SURFACE OF EARTH HAVING LATITUDE π/2 . ITS MASS AT CENTRE OF EARTH WILL BE (kg)

 

1) 10/ 37√2

2 ) 10√2

3) 5

4) 15

Asked By: SARIKA SHARMA
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Joshi sir comment

π/2 means it is pole so mg = 9 implies that m = 9/9.8

same will be the mass at centre of earth

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

a double star consist of 2 stars of mass m & .2m. distance b/w them is r. d 2 stars revolve under their mutual gravitatn force then d 2 stars have equal periods of revolution . HOW?

Asked By: SARIKA SHARMA
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Joshi sir comment

according to newtons third law both star will give equal force to each other

so for first star of mass m,           m(2r/3)ω2  = F or ω = [3F/2mr]1/2

similarly for second                   2m(r/3)ω2 = F   or  ω = [3F/2mr]1/2

here 2r/3 and r/3 are taken according to centre of mass criteria.

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

WEIGHT OF A BODY ON SURFACE OF EARTH DEPENDS UPON D ANGULAR SPEED OF ROTATn OF EARTH ABOUT ITS OWN AXIS. IS THIS STATMENT IS CORRECT? PLS. GIVE REASON ALSO

Asked By: SARIKA
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Joshi sir comment

since a cetrifugal force works on the body and this force depends on angular speed so statement is correct.

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

Q IF A BODY IS PROJECTED UPWARD WITH SPEED 1/2 OF ESCAPE SPEED OF EARTH THEN MAXn HEIGHT ATTAINED BY IT IS=? 

ANS=R/3  WHERE R=RADIUS OF EARTH

Asked By: SARIKA Solved By: NIKHIL VARSHNEY
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Joshi sir comment

escape velocity = √(2GM/R)   so 1/2 of escape = 1/2√(2GM/R)

so according to given condition on applying energy conservation we will get 

energy at start = energy at the heighest point

or    -GMm/R +  mv2/2 =  -GMm/R1 +  0

or     - GMm/R   +  2GMm/8R =  -GMm/R1 + 0

solve now for finding R1  then find R1-R

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

Q A UNIFORM SPHERICAL SHELL GRADUALLY SHRINKS MAINTAING ITS SHAPE. THE GRAVITATn POTENTIAL AT THE CENTRE WILL DECREASE. Y DOESN'T IT SHD REMAIN CONST. =?

Asked By: SARIKA
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Joshi sir comment

Formula for gravitational potential at centre is -3/2 GM/R so if R will decrease V will increse numerically but - is present so decreases

please,  write tibs about tidal effect  and it's fourmula ?   Is it depend inversly on the cube of the distance? and is  tidal effect of the moon's pull grater than the tidal effect of Sun?

 

Asked By: SHUBHAM VED
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Joshi sir comment

 

The tidal force is a secondary effect of the force of gravity and is responsible for the tides. It arises because the gravitational force per unit mass exerted on one body by a second body is not constant across its diameter, the side nearest to the second being more attracted by it than the side farther away. Stated differently, the tidal force is a differential force. Consider three things being pulled by the moon: the oceans nearest the moon, the solid earth, and the oceans farthest from the moon. The moon pulls on the solid earth, but it pulls harder on the near oceans, so they approach the moon more causing a high tide; and the moon pulls least of all on the far oceans (on the other side of the planet), so they stay behind more, causing another high tide at the same time. If we imagine looking at the Earth from space, we see that the whole Earth was pulled, but the near oceans more and the far oceans less; the far oceans stayed behind since they are pulled less (since they are farther away).

 

Tidal effects become particularly pronounced near small bodies of high mass, such as neutron stars or black holes, where they are responsible for the "spaghettification" of infalling matter. Tidal forces create the oceanic tide of Earth's oceans, where the attracting bodies are the Moon and, to a lesser extent, the Sun.

 

Formula

\vec a_t(axial)  ~ \approx ~ \pm ~ \hat r ~ 2 \Delta r ~ G ~ \frac{M}{R^3}

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