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Gravitation

The document contains a series of physics problems related to gravitational forces, satellite motion, and potential energy. It includes calculations for gravitational acceleration at various depths, the effects of distance on gravitational attraction, and the behavior of objects in circular orbits. Additionally, it discusses the implications of changes in planetary rotation and mass on gravitational forces.

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Aditya Jadhav
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0% found this document useful (0 votes)
36 views3 pages

Gravitation

The document contains a series of physics problems related to gravitational forces, satellite motion, and potential energy. It includes calculations for gravitational acceleration at various depths, the effects of distance on gravitational attraction, and the behavior of objects in circular orbits. Additionally, it discusses the implications of changes in planetary rotation and mass on gravitational forces.

Uploaded by

Aditya Jadhav
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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6.

If the value of gravitational acceleration at a


1. Consider a light planet revolving around a depth d below the surface of the earth is twice to
massive slar in a cirular orbit ot radius r'
that on the surface of the earth at latitude of
with time period 'T, Ifthe gravitational torce angle . then
ot attraction between the planet and the star is Ro
(A) =cos
proportional to r.then T is proportional to gd
(A) n )2 (B)
() (B) =cos
(D) R´o*

2. If the distance between the sun and the earth is


(C) =cos - Vg(R +d)
increased by two and a half tinmes, then V2 Ro
attraction between two wvill
Vgd
(D) = cos|
(A) rema constant. (B) decrease by 63%. Ro
(C) decrease by 84%. (D) decrease by 890 0.

7. If the earth of radius R. while rotating with


3. The correet graph representing the variation of angular velocity o becomes stand still, what
total energy (T.E.). kinetic energy (K.E.) and will be the ettect on the weight of a body of
potential energy (P.E.) of a satellite with its mass m at latitude of 60°
distance (r) from the centre of the earth is (A) Remains unchanged
(B) Decreases by R o
(A) (B)
(C) Inereases by R o/4
(D) Increases by Ro2.
T.E
K.E. 8. Time period of second pendulum on a planet.
P.E. + whose mass and diameter are half that of earth,
K.E. is
P.E.
T.E (A) V2s (B) 2s
(C) 2V2 s (D)
(C) (D)
E 9. A body of mass m rises to a height h = R/6 from
the surtace of the Earth. If g is acceleration due to
T.E
gravity on the Earth's surtace. then the inerease
K.E. K.E. in potential energy is (R=radius of Earth)
T.E P.E. (A) mgh (B)
4
mgh
P.E.

(C) mgh (D) mgh


4. A body vweighs S0N on the surtace of the Earth
while at adepth d below the surface Earth, the 10. Two-point masses each equal to 1.7 kg attract
weight is 25 N. The value ofd is one another with a force of 2.89 x 10 kg-Wt.
(A) 0.4 R (B) 0.5 R The distance between the point masses is
() 0.8 R (D) R (A) 8 cm (B) 0.8 cm
(C) 80 cm (D) 0.08 cm
5. Time period of revolution of a nearest satellite
11. Consider a planet in some solar system which
around a planet of radius R is T. Period of
revolution around another planet, whose radius has mass half the mass of the Earth and density
is SR but density twice the first planet is equal to the average density of the Earth. An
(A) T/ V2 (B) 3T
object weighing Won the Earth willweigh
(A) 23w (B) 2W
(C) ST (D) 5/3T (C) W/2 (D)
12. Asatellite is launched into a circular orbit of 20. The gravitational potential due to a point mass
radius R around the earth. A second satellite is Mat a distance r (>R) from the centre of earth is
launched into an orbit of radius (1.02)R. The given by,
period of the second satellite is larger than that -GM V=8
(A) V= (B)
of the first one by approximately. r

(A) 0.5% (B) 3.0% - GM


(C) V=8 (D) V=
(C) 1.5% (D) 1.0%
21. The gravitational force on a body of mass 8kg
13. The acceleration due to gravity at a height 1/10 at the surface of the earth is 80 N. If earth is a
of the raius of a spherical planet above the its perfect sphere, the gravitational force on a
surface is 10.5 ms. Its value at a point at an satellite of mass 350 kg in a circular orbit of
equal distance below the surface in ms - is about radius same as diameter of the earth is
(A) 8.5 (B) 9.8 (A) 80 N (B) 350 N
(C) 11.81 (D) 13.12 (C) 3500 N (D) 875 N

14. Two equal masses, each equal to m are 22. A spring balance is graduated on sea level. If a
suspended from a balance whose scale pans body is weighed with this balance
differ in vertical height by h. The error in consecutively increasing depth from earth's
weighing in terms of density of earthp is surface, the weight indicated by the balance
(A) will go on increasing continuously.
(A) nGpmh (B) Gpmh (B) willgo on decreasing continuously.
(C) will remain same.
8 4
(C) nGpmh (D) 3 Gpmh (D) will first increase and then decrease.
23. [M'L'Tare dimensions of
15. The maximum vertical distance through which a (A) gravitational torque.
full dressed astronaut can jump on the earth is (B) gravitational potential
60 cm. Estimate the maximum vertical distance (C) universal gravitational constant
through which he can jump on the moon, which (D) gravitational field
has a mean density 2/3 that of earth and radius
one quarter that of the earth. 24. To increase gravitational acceleration of the
(A) 1.5 m (B) 3.6 m planet
(C) 6m (D) 7.5 m (A) its size should decrease with its mass
unchanged.
16. Mass M is divided into two parts xm and (B) its mass should increase with its size
(l- x)m. For agiven separation, the value of xm unchanged.
for which the gravitational attraction between (C) its mean density should increase.
the two pieces becomes maximum is (D) all of the above.
3 25.
(A) (B) (C) (D) 2 The value of 'g' at a certain heighth above the
5
X
free surface of Earth is where x is the value of
17. 8
The value of acceleration due to gravity at a
depth of 800 km is equal to 'g'at the surface of Earth. The height h is
(A) 8.58 ms (B) 9,8 ms (A) 0.8R (B) 1.2R
2 (C) 1.83R (D) 8R
(C) 7.35 ms (D) 19.6 ms
26. If earth stops rotating then as a consequence,
18. A satellite is orbiting close to the earth and has a
kinetic energy E. The minimum extra kinetic (A) centripetal force will vanish and value of
'g' will increase.
energy required by it to just overcome the centrifugal force will vanish and value of
(B)
gravitation pullof the earth is 'g' will decrease.
(A) E (B) VE (C) value of 'g' will become zero.
(C) V2 E (D) 2E (D) day and night will not occur but value of
'g' will be unaffected.
19. Two satellites A and B go round a planet in
circular orbits having radii 9R and R., 27. A body weighs 27 N on the surface of the earth.
respectively. If the speed of satellite A is nv, What is the gravitational force on it due to earth
then speed of satellite B is at a height equal to half the radius of the earth
(A) 9 ny from the surface?
(B) 3 ny (A) 72 N (B) 27 N
(C) V3 ny (D) (C) 36 N (D) 12 N
34. 33. 32. 31. 30. 29. 28.

(allyD) (C) (B) periodic (A) surface


The
time is A (C) surface.
(A) satellite The (C) (Range (A)
A) of (C) between Two
The 2r, solid (D) of A (C) velocity
(C) (B) (satellite isangular
(A) earth Ifsunthe inThe
A) andswept (A)87
(D) (C) minutes. distance
(B) surface surface The
willbeperiod
made satellite
mass gravitational
value mutual earth rotational
density
planet.
of
inverselyplanet.of
directly
directly 8000km 12500km infinite
light-year
100 proportional m.
proportional inversely
to directly
directly L.
to directly by 4m/s2x10 is minutes.
minutes.
261 452 87
of them momentum mE x three
of of
gravitational
force is of the 10
of3.6
circular(mass x another
is the of of revolution V3 the
orbiting height same line in
proportional
proportional earth gravitational proportional
proportional mass x times
g is proportional proportional a rad/s minutes. earthperiod
oportional proportional spheres joining circular
force =6x
at material,
keptare 10 orbit earth
is is a m the
very L. exerted
is is. of
25%
(R= certain moving N, of10*
6400(D)km(B) (D) (B) The (D) (B ) satellite
radius 87 a
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to to close 6400 of force to the orbit kg) satellite
density to to
to to centre
L by minutes.
tosquare
10500km height 1000 finite rate around of
ity to
ofsquare to km) its parsec L
m Land of 2 4earth's revolves in
inversely
and and and x x the 1.5
a value of radius
of 10 10
earth an
close
contact.in m. m. of sun x
of ofplanet. above attraction orbit The
root densityplanet. inversely the the the rad/s m/s
10* from
on R angular onaround to
and earth area has earth km. time
at
Its the the the
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40. 39. 38. 37. 36. 35.

(C) g-wt(978A) earth?theof earth. A (D) (C) (B) (earth's then


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(A) massradiusThe ofKepler' Cavendish determined
(D) balance.
(sC) balance.
(B)spring(A) The gravitational
(C)6R(A) the the ofand Mass (C) (A) Thenradius earth V
planet a body the the andV,
distance
line distance
g-wt
1120 Decreases
How Decreases
Increases by3%
1.5%Increases
by3% by what earth -2.5%
remaining same. effect. value 30R of Ve
whose weighs inthe Compton =Vp Ve and and
by joining planet. a =
the planet 2Vp another
much surface
now? willsuddenly
1.5% using between the denote
value force between
mass 1400 by happen Earththe pendulum. gravitationalof If same
will the R is
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is contracts moon is 100
g-wt
it g the the then mean escape
(D) (B) weigh mass
to +5%(D) (B) is (D) (B) moon times
and shrinks (D) (B) having
on the zero and radius 50 is density
radius onthe remaining so +2.5% percentageThe Ve Ve
velocities
g-wt670 1400
g-wt escape 4.55R 15R is, the and times mass
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