JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE

SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
1.
Ans. (A)
Sol. To avoid disintegration
GM
2
³ w2 r
r

Gæ4 3 ö
2 ç
pr r ÷ ³ rw2

®
r è3 ø

4pGr
w£
3

2p 4pGr
£
T 3

3p
T³
Gr
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
2.
Ans. (A)
Sol. Conserving angular momentum, we have
mvR = mv'r ... (1)
[v' = speed when spaceship is just touching the planet]
Conserving energy, we have
1 1 GMm
mv 2 = mv '2 - ... (2)

®
2 2 r
1/ 2
r é 2 2GM ù
Solving (1) & (B), we get, R = v +
v êë r úû
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
3.
Ans. (A)

Sol. h

®
Guass law for gravitation
r r
ò g · d s = – min. 4pG
GM
g=
R2

GM M
2× ×A = (h × A) × 4pG
R2 4 3
pR
3

2R
Þh=
3
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
4.
Ans. (A,B,C)
Sol. For (A) : By COME total mechanical energy of the two objects
For (B) : By using reduced mass concept

1 2 G(4m)(m) (m)(4m) 4 10Gm

µv rel – = 0 where µ = = m Þ vrel =
2 r m + 4m 5 r

®
G(m)(4m) 4Gm 2
For (C) total KE = –PE = =
r r
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
5.
Ans. (A,D)

V
Sol.
R/3

®
Total potential energy at Pt 'P'

æ GMm é 3 ( 2 / 3R )2 ù 3GMm ö
ç ê - ú- ÷
ç R ê2 2R 2
úû 2 ´ R / 3 ÷
è ë ø

20GMm 1
= = mv 2p
18R 2

20GM
vP =
9R
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
6.
Ans. 3
1
Sol. w.r.t. COM K.E. = (red mass) vrel2
2

mr
w.r.t. COM Angular momentum = v
2 rel
\ Equating energy

®
1 m 2 Gm 2 1 m 2 Gm 2
v0 - 0 = vrel -
2 2 r 2 2 r
(Here vrel is relative velocity | to line as vrel along the line joining is zero when separation is either min.
or max.)
Angular momentum conservation
mr0 mr
v0 = v
2 2 rel

r0 v0
\ vrel =
r
solving 3r2 – 4rr0 + r02 = 0
r0
\ rmax = r0 rmin =
3
ratio = 3 (ans)]
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
7.
Ans. 2.01 sec

l
Sol. T = 2p = 2 sec.
g

1/ 2 1/ 2
l l æ xö
= 2 ´ æç1 -
64 ö
T' = 2p = 2p = 2 × ç1 - ÷ ÷
g¢ æ xö è Rø è 6400 ø

®
g ç1 - ÷
è Rø

æ 1 ö
» 2 ´ ç1 - ÷ » 2.01 sec
è 200 ø
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
8.
Ans. 99.5R

B
h

R/100
Sol. A

®
Let particle be projected from A and it reaches to point B at a height h from surface.

2GM
It is projected with escape velocity =
R
Energy at A = energy at B
K.E. + P.E. (at A) = P.E. at B

1 2GM GmM é 2 æ 99 ö 2 ù
2
GMm
m - 3 ê
3R - ç ÷ R ú=-
2 R 2R êë è 100 ø úû (R + h )

1 2GM 2.0199GmM GMm

m - =-
2 R 2R (R + h)
0.0199GMm GMm
- =-
2R (R + h)
2r
R+h = = 100.5R
0.0199
h = 99.5 R
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
9. GR0009
Ans. 2R, 3R[3 – 3

3M12R (M,R)

Sol. m

®
(a) speed will maximum at projection point
(b) To just reach the other planet small body must be able to reach the point where gravitation field due
to both planet is zero. At this point speed will be minimum and equal to zero.
G ´ 3M GM
= Þ 3 ( 6R - x ) = x
( 6R - x )
2 2
x

(
Þ 1 + 3 x = 6 3R )

x=
6 3 ( )
3 -1 R
2
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
10. GR0012
Ans. T = 3 × 10–2 N
Sol. T + Fg = mar
GMm
T+ = mw2(R+ l )
( R + l )2

-2
GMm æ lö
T = mw R + l ) -
2(
2 ç
1+ ÷

®
R è Rø
-2
æ lö æ lö é gù
êëQ w = R úû
2
T = mg ç1 + ÷ - mg ç1 + ÷
è Rø è Rø

3mgl
T=
R
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
11. GR0041
Ans. (A,C,D)

Sol.
A(–2,0,0) B(2,0,0)
x
O

®
z

Let
r
FA = gravitational field due to sphere A (if it were present at that position)
r
FB = gravitational field due to sphere B (if it were present at that position)
r
FR = gravitational field due to remaining portion after the cavities are made.
r r r
Then from superposition principle, we can see that FA + FB + FR = 0, as the force due to entire, sphere is
zero at centre.
r r
Now since FA + FB = 0 due to symmetry.
r
Hence FR = 0, hence option (i) is correct
Now at B.
r GM GM GM
Field due to entire sphere is given by F = 3 r = 2= .
R 64 32
r r r
Then from superposition principle, we can see that FA + FB + FR = 0 , as the force due to entire sphere is
zero at centre.
r r
Now since FA + FB = 0 due to symmetry.
r
Hence FR = 0 , hence option (i) is correct
Now at B,
r GM GM GM
Field due to entire sphere is given by F = 3 r = 2= .
R 64 32
r Gm Gm GM GM r
whereas FA = 2 r = = = , where m = mass of sphere A = M/64 and FB = 0
4 16 16 ´ 64 1024
From superposition principle, we have
r r r r
FA + FB + FR = F
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
r r r GM ˆ GM ˆ 31GM ˆ
Þ FR = F - FA = i- i= i ¹0,
32 1024 1024
Hence option (ii) is not correct.
Regarding potential at a point on y2 + z2 = 36, we can see that the radius of the circle is 6 units, howerver,
all the points on it are symmetrically located from remaining sphere. Hence potential must be same at
every point on this circle. Same logic holds for the circle y2 + z2 = 4 also, though this circle lies inside the
remaining sphere.

®
Hence options (iii) and (iv) are also correct.
Hence options (i), (iii) and (iv) are corect.
Note, we can use superposition principle to calculate the potential at these points.

GM æ 1 1 ö
In option (iii) it will be equal to - -
2 è 3 32 10 ÷ø
ç

GM æ 1 ö
In option (iv) it will be equal to - ç 22 - ÷
64 è 2ø
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
12.
A ns. (a) U = – Gm1m2/r ; (b) U = –G(mM/l) ln (1 + l/a) ; F = GmM/a(a + l)

-GMm
Sol. U = M
r M
r
-Gmdm
dU =
x
x dx

®
m M
( a +h )
-GmM dx
ò dU = l ò x a
a

-GmM æ a + l ö
U= ln ç ÷
l è a ø

a +l
GMm dx
F=
l ò x2
a

GMm
F=
l (a + l)
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
13.
Ans. (A)
Sol. For circular orbit

GM
v= .....(i)
2R

1 GMm GMm
mv 20 - =-

®
2 2R 3R

v 20 GM æ 1 1 ö GM
= ç - ÷ Þ v 20 =
2 R è2 3ø 3R

GM
3v 20 = ....(ii)
R

3
v= v0
2
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
14.
Ans. (A)
Sol. q = w1t
q + p = w2t
p = (w2 – w1)t
p
t=
w2 - w1

®
GM
& w=
r3
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
15.
A ns. (B)

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
16.
Ans. (D)

rA rB
mA mB
Sol. C

®
Gm A m B m A rA 4p2 m B rB 4p2
= =
( rA + rB )
2
TA2 TB2

Þ mArA = mBrB
\ TA = TB
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
17.
Ans. (B,C,D)
Sol. It is a binary star system, so, they have

G ( m1 + m 2 )
w=
( R1 + R 2 )
3

Same for both,

®
hence, equal time period of 2 days
For A, T = 2 days
2pR 1
= T = 2days
V1

R1 =
( 2days ) V1 ( 2 ´ 24 ´ 3600s )( 3.64 ´10 )
4

= = 1 × 109
2p 2p

( R1 + R 2 )
3
2p
T= = 2p
w G ( m1 + m 2 )

4p2 ( R 1 + R 2 )
3

Þ ( m1 + m 2 ) = 2 = 5.35 × 10+29 kg.

T G
and m1w2r1 = FG = m2w2r2
m1 r2
Þ m1r1 = m2r2 Þ m = r = 2
2 1

Þ m1 = 3.57 × 1029 kg & m2 = 1.78 × 1029 kg

 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
18.
Ans. (A,B,C,D)

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
19.
Ans. (C)
Sol. T µ n3

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
20.
Ans. (B)
Sol. R µ n2

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
21.
Ans. (C)
Sol. Both planets have same mass and same length of day. Geostationary satellite - planet system will have
same energy in either planet. But the satellite - sun system will account for the energy difference.

GMm0 GMm0
Ui = – + Usatellite – planet Uf = – + Usatellite – planet
2r 2 (4r )

®
3 GMm0
Emin = Uf – Ui =
8r
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
22.

2p
Ans.
é GM ù
3ê 3
- we ú
ë 8R û
Sol. Let the angular speed of the satellite is w.
GMm

®
So, 2
= mw2 ( 2R )
( 2R )

GM
w=
R3
Relative angular velocity is

GM
wr = - we
8R 3
Angle subtended at the horizon is

æqö 2p
cos ç ÷ = 0.5 Þ q =
è2ø 3

q 2h
T= ÞT=
wr é GM ù
So, 3ê - we ú
3
ë 8R û
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
23.
Ans. 2GMm 2 b
Sol. When particle is morning in a parabolic orbit i.e. its velocity of the particle is equal to escape velocity.
So, Angular momentum is,

2GM
L =m ´ b Þ L = 2GMm 2 b
b

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
24.

GMm 4pfr 3
Ans. (a) – , (b) , (c) DK = +2prf
2r0 GMm
Sol. (a) TE = PE + KE
- GMm 1 4Mm
= +
r 2 r

®
- 2 GMm + GMm - GMm
= =
2r 2r

d ( TE ) GMm
(b) =– (–1)
dr 2r 2

dTE GMm
=
dr 2r 2
d(TE) = work done by friction = f × 2pr

f 4pr 3
dr =
GMm
(c) DKE + Dwnet = 0
DKE = 2prf
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
25. GR0010

æ 7 ö
A ns (a) h = çç +1÷÷ R, (b) 1.125 R
è 2 ø

V0
a

®
Sol. (a)
u

Apply consentration of angular momentum.

Li = 2f
mv0cosaR = mur
v 0 cos aR
u= ____(1)
r
Apply consentration of Energy,

æ 7ö
In solving r = ç 2 + ÷R
è 2 ø

æ 7ö
So, h = r – R = ç1 + ÷R
è 2 ø
(b) Radius of curvature of trajectory is :

u 2 v 02 cosa2 R 2
r= =
g r2 ´ g
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
26. GR0013
1 2
Ans. K Gp(R4 – r4)
4
Sol. dm = kr4pr2dr
m = kpr4
Gm
E=
r2

®
dr
r
–dP × 4pr2 = G m dm
r2

Gkpr 4 ´ k4 pr 3dr
-dP =
4 pr 4
–dP = Gk2pr3dr
P r= r 3
-ò dP = Gk 2 p ò r dr
P= 0 r= R

–P = Gk p
( r4 - R4 )
2
4

R4 - r 4
P = Gpk 2
4
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
27. GR0015
Ans. 3

GM1 M 2 M 2 M1w S2 ( 4r )
Sol. =
( 4r ) 2 ( M1 + M2 )
2
æ vö GM
w =ç ÷ = 3 S
2
eè rø r

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
28. GR0017

æ x2 - R2 ö
Ans. ç1 - ÷4pR 2
ç x ÷
è ø

®
R
q
Sol. x

Area = 2 (1 – cos q) 2pR2

é x2 - R2 ù
Area = 4pR ê1 -
2
ú
ëê x úû
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
29. GR0043
Ans. (C)

dA r 2 æ dq ö
Sol. = ç ÷
dt 2 è dt ø

æ dq ö 2 æ dA ö
w=ç ÷ = 2 ç ÷
è dt ø r è dt ø

®
and v = rw
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
30. GR0044
Ans. (A)
Sol. Let MG & ms represent masses of the galaxy and the Sun.
3 2
4p 2 3 4p 2 3 M æRö æ t ö
A ccording to K epler's law t = r and T 2 = R Þ G =ç ÷ ç ÷ n
2

Gms GM G ms è r ø è T ø

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
31. GR0049
Ans. (A,C,D)
Sol. There is no external torque is acting in any case.

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
32.
Ans. 64.5 days
Sol. Falling of the body on the Sun can be considered as the motion along a very elongated (in the limit,
degenerated) ellipse whose major semi-axis is partically equal to the radius R of the Earth's orbit. Then
from Kepler's laws, (2t/T)2 = [(R/2)/R]3, where t is the falling time (the time needed to complete half a
revolution along the elongated ellipse), T is the period of the Earth's revolution around the Sun. Hence
t = T/4 2 = 65 days

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
33.
r0 é1 ± 1 - (2 - h) h sin 2 a ù
Ans. rm = 2 - h úû , where h = r0v0 /GmS, mS being the mass of the sun.
2
êë

Sol. From the conservation of angular momentum about the Sun,

mv0r0sin a = mv1r1 = mv2r2 or, v1r1 = v2r2 = v0r0 sin a (1)
From conservation of mechanical energy,
1 GM s M 1 GM s M

®
mv02 - = Mv12 - ______(1)
2 r0 2 r1
Solving equation (1) and equation (2)

æ 2 gm s ö
( )
-2 gm s ± 4 g 2 m s2 + 4 v 20 r02 sin 2 a ç v20 -
r0 ø
÷
So, r1 = è
æ 2 gm s ö
2 ç v 20 - ÷
è r0 ø

v20 r02 sin 2 a æ 2 v 02 ö

1± 1- ç - ÷
gm s r0 é1 ± 1 - ( 2 - h) h sin 2 a ù
è r0 gm s ø
= = ë û
æ 2 v20 ö ( 2 - h )
ç - ÷
è r0 gm s ø

v20 r0
where h = , (ms is the mass of the Sun).
gm s
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
34.
Ans. 6.27 × 1024 kg
Sol. M = (4p2R3/gT2) (1 + T/t)2 = 6.1024 kg, where T is the period of revolution of the earth about its own
axis.

®
 JEE (MAIN + ADVANCED) : ENTHUSIAST COURSE
SUBJECT : PHYSICS
Topic : Student Question Bank (Gravitation)
35.

m ( )
h -1
Ans. t =
a gR

Sol. The decrease in the total energy E of the satallite over the time interval dt is equal to –dE = Fv dt.
Representating E and v as functions of the distance r between the satellite and the centre of the Moon,

( ) m/a

®
we can reduce this equation to the form convenient for integration. Finally we get t » h -1 gR