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2. If three particles each of mass M are placed at the three
comers of an equilateral triangle of side a, the forces
exerted by this system on another particle of mass M placed
(i) at the mid-point of a side and (ii) at the centre of the
triangle are respectively:
2
oon
(b) 4GM
3a?
2 2
(¢) 3GM~ GM" (4) 0,0
a2 a
2
30 ) Gravitational force on the particle
placed at the mid- ‘Point D of side AB of
on ais= A + BR + B. But R and
A are equal in magnitude and opposite
in direction.
2
F=Fy= 4GMm _ 4GM
“3a 3a
cp? -32@
f-cot-4]
(i) Gravitational force on the particle Mac
placed at the point of intersection of
three medians =R+R%+ R=0.
Since, the resultant of F and 5 is M 7
equal and opposite to B. A B » gopondence of acceleration due to gravity g on the
om ce r from the centre of the earth, assumed to be a
oe of radius R of uniform density is as shown in figures
oes below. The correct figure is:
[AIPMT 2010 ; NEET 2016 (Phase-I}]
() 9 (2) 9
«
pl------=
GB) 9 (4) 9
xp h------>>
n}------ 6. The acceleration due to gravity at a depth d below surface of the
earth is,
ee t-a)=«(-9)
g'=Oatd=R
i.e. , acceleration due to gravity is zero at the centre of the earth.
Thus, the variation in value of g with ris:
Forr>R:
1 & gk?
oT ae
l+5 r
(3)
or grat
Pr
Here, R+h=r
: gtag(1-2)-&
Forr or is or x
"alse
ON coca lena re recreate reg Ny ARR 7 RE py RTT 28.
wr very wean
Two bodies with masses M and. M? are initially at rest and
a distance R apart. They then move directly towards one
another under the influence of their mutual gravitational
attraction. What is the ratio of the distances travelled by M,
to the distance travelled by M2?
M, M> 1
— b) —+ —
(a) M> (b) M (c) 1 (d) 2 eee ye
28. Because gravitational force is the mutual force, hence position 4
the centre of mass remains unaffected. 7
M 1 =M: ‘R2
R _ M2
* RM" 29. Ifg is acceleration due to gravity on the earth’s surface, the
gain in the potential energy of an object of mass m raised
from the surface of the earth to a height equal to the radius 2
of the earth is: [AIIMS 2000; Kerala (PMT) 2006;
DPMT 2006; MPPMT 2007; NEET 2013]
(2) jmeR (b) 2mgR
(c) mgR (d) } mgR 29, Ata distance x, ., from the centre of the earth, gravitational force,
F= cMm
x
aW = Fdx = 2M gy
x
_pRth GMm oy [14
w=f, 3 ar=Gun|2-
21 1 .
=ak'm [ eal 0
-. Gain in potential energy
= gR2m| 41-1]. ms
aR'n| 3 vas .
anor ay vee ey tee
38. A tunnel is dug along a diameter of the earth of mass M,
and radius R,. The force on a particle of mass m placed in
the tunnel at a distance r from the centre is:
[AMU (Med.) 2006]
@) eee r (b) GM,.m
R3 Ror
3
©) GM ,mRe @) GM..m r
r R? 38. uv
40. A satellite of mass m revolves around the earth of radius R
at a height x from its surface. If g is the acceleration due to
gravity on the surface of the earth, the orbital speed of the
satellite is:
gR
(a) gx (b) =
2 /2
@ | se
R+x R+
x ’
40. For the satellite, the gravitational force provides the necessary
centripetal force, i.e. ,
GM,m m2 , GM,
— = d e-
(Rex Rx) oR? 8
2
aoa x. 47.
If three particles each of mass M are placed at the corners of
an equilateral triangle of side a, the potential energy of the
system and the work done if the side of the triangle is
changed from a to 2a, are:
(@) 36M 36M () 36M? 3GM?
ae’ 2 a” 2a
(c) 36M? 36M? ( 36M 36M
a’ 4a2 a ° 2a L ~ 4
4. U4 = Potential energy of the system
_ GMM _GMM _GMM
a a a
_3GM?
a
mt __GMM _GMM _GMM
a Us 2a qij2aes 128
__ 36M?
2a
Work done = change in potential energy
36M?
HUB UAH IQ
Avow hetweell Masses 52. If a particle is fired vertically upwards from the surface of
the earth and reaches a height of 6400 km, the initial
velocity of the particle is: (Assume R=6400km and
g=10m s?)
(a) 4 km/sec (b) 2 km/sec
(c) 8 km/sec (d) 16 km/sec $2. According to law of conservation of energy,
1 yy? = meh
2 144
R
2_ 2gh _2x10x64 x10°
rh WGRk
Its 1+s
_2x10x64 «109
2
v= 764 x10° =8 km/sec.
.
