Solution

WORKSHEET

Class 11 - Physics
1. A radius of gyration in general is the distance from the center of mass of a body at which the whole mass could be concentrated
without changing its moment of rotational inertia about an axis through the center of mass.
since mn = M total mass of the body,
2 2 2
M(r + r +⋯+ rn )
1 2
I =
n

From the above equations, we have

2 2 2 2
MRg =M(r + r +⋯+ rn )
1 2

Radius of gyration is the root mean square distance of particles from axis formula
2 2 2 2
R =(r + r +⋯+ rn )
g 1 2

Therefore, the radius of gyration of a body about a given axis may also be defined as the root mean square distance of the various
particles of the body from the axis of rotation.
2. “Rotational motion can be defined as the motion of an object around a circular path, in a fixed orbit.” The dynamics for rotational
motion are completely analogous to linear or translational dynamics.
For rotational motion,
Angular acceleration, α = dω

dt

dω = α dt
Integrating both sides within limits.
ω
f t

∫ dω = ∫ αdt
ωi 0

If α is constant,
ωf t
[ω ]ω = α[t]
i 0

ωf − ωi = αt ...(i) At any instant, ω = (ω 1 + αt )

dθ
Angular velocity, ω = dt

θ2

∫ dθ = ∫ ωdt
θ1

t
θ2
[θ] = ∫ (ωi + αt) dt
Θ1
0
2

(θ2 − θ1 ) = ωi t +
αt

2
...(ii)
Also, α = ω dω

dθ

ωi θ ω
2 f
ω θ
∫ ωdω = ∫ αdθ ⇒ [ ] = α[θ]
2 0
ωc
ωj 0

2 2
ω −ω
f i
= 2θ
2

2 2
ω − ω = 2αθ
f i

3. Let M be the mass and Lbe the length of the metre scale. When the upper end of the rod strikes the floor, its centre of gravity falls
through height . L

M.I. of the scale about the lower end A, I = M.I. of the scale about the parallel axis through CG + Md2
2 2 2
2 ML ML ML L
= I0 + M d = + = [∵ d = ]
12 4 3 2

Also, ω = v

r
=
v

Gain in rotational K.E.

2 2 2
1 2 1 ML v Mv
= Iω = ⋅ ⋅ =
2 2 3 2 6
L

Now, Gain in rotational K.E. = Loss in P.E.

1 / 13
 2
Mv

6
= Mg ⋅
L

2
or v
2
= 3gl

−−− −−−−− −−−− −1

or v = √3gL = √3 × 9.8 × 1 = 5.4 ms

4. A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, v = 5 m/s

Angle of inclination, θ = 30 ∘

Height reached by the cylinder = h

a. Energy of the cylinder at point A will be purely kinetic due to the rotation and translational motion. Hence, total energy at A
= KErot + KEtrans
1 1
= 2
Iω
2
+
2
mv
2

The energy of the cylinder at point B will be purely in the form of gravitational potential energy = mgh
Using the law of conservation of energy, we can write:
1 2 1 2
Iω + mv = mgh
2 2

Moment of inertia of the solid cylinder, I = 1

2
2
mr

1 1 2 2 1 2
∴ ( mr ) ω + mv = mgh
2 2 2
1 2 1 2
Iω + mv = mgh
4 2

But we have the relation, v = rω

1 2 1 2
∴ v + v = gh
4 2
3 2
v = gh
4
2
3 v
∴ h =
4 g

3 5×5
= × = 1.91m
4 9.8

To find the distance covered along the inclined plane

In ΔABC :
BC
sin θ =
AB

∘ h
sin 30 =
AB
1.91
AB = = 3.82m
0.5

Hence, the cylinder will travel 3.82 m up the inclined plane.

1

2
⎛ 2gh
⎞
b. v = K2
⎝ 1+ ⎠
2
R
1

2
⎛ 2gAB sin θ
⎞
∴ v =
2
K
⎝ 1+ ⎠
2
R
2

For the solid cylinder, K 2

=
R

2
1

2
2gAB sin θ
∴ v = ( )
1
1+
2

4 2
= ( gAB sin θ)
3

The time taken to return to the bottom is:

AB
t =
v
1

AB 3AB 2
= = ( )
1 4g sin θ
4
( gAB sin θ) 2
3
1

11.46 2
= ( ) = 0.7645
19.6

So the total time taken by the cylinder to return to the bottom is (2 × 0.764)= 1.53 s.as time of ascend is equal to time of
descend for the following problem.
–
5. (a) 1 : 2√2
Explanation:

2 / 13
 −−−−−
−−−
− 8
ve = √2gR = R√ πGp
3

νs R√ρ

∴ =
vp
Rp √Pp

R ρ
√
= =
1

2R × √2p 2√2

–
= 1 : 2√2
6.
(b) 2 E0
Explanation:
GMm
Total energy = − 2r
= E0

Potential energy = − GMm

r
= 2E0

7.
(d) -4G
Explanation:
G × 2 G × 2 G × 2 G × 2
V=− 1
−
2
−
4
−
8

= −2G [1 + 1

2
+
1

2
+
1

3
+ …]
2 2

= −2G 1

1
= -4G
(1− )
2

8.
(b) h = R
Explanation:
At h = R the acceleration due to gravity would be one forth of its value on earth surface.

9.
(d) 1

Explanation:
CMm
P.E. = − r
2
GMm mv
Now 2
=
r
r
1 2 GMm
∴ K. E. = mv =
2 2r
KE GMm r 1
= × =
P.E 2r GMm 2

10.
(d) acceleration
Explanation:
Acceleration due to gravity is independent of the mass of the body.

11.
(d) 7460 m/s
Explanation:
Mass of the Earth, Me = 6.0 × 10 kg 24

The radius of the Earth, Re = 6.4 × 10 m 6

Universal gravitational constant, G = 6.67 × 10 −11 2

N m kg
−2

Height of the satellite, h = 780 km = 780 × 103m = 0.78 × 106m

−−−−
GMe
Orbital velocity of the satellite, v = √ Re +h
−−−−−−−−−−−−−
−11 24
6.67× 10 ×6.0× 10
= √
6 6
6.4× 10 +0.78× 10

−−−−−−
13
40×10
= √
6
7.18×10

3 / 13
 −−−−−−− −
7
= √5.57 × 10

3 −−−−−−−−
= 10 × √5.57 × 10

3
= 7.46 × 10

= 7460 m/sec

12. The field of force surrounding a body of finite mass in which another body would experience an attractive force that is
proportional to the product of the masses and inversely proportional to the square of the distance between them.
Gravitational field intensity is the force of gravitation per unit mass
F GMm GM
I = = =
m R2 2
R
m
GM
Also, g = 2
R

So, Acceleration due to gravity is same as the gravitational field intensity.

13. Orbital velocity is the velocity at which a body revolves around the other body. Objects that travel in the uniform circular motion
around the Earth are called to be in orbit.
For circular orbit
2
mV

F=
0

r
2
GMm mv
⇒ =
2 r
r
−−−
GM
⇒ V0 = √
r

Now for satellite very new to earth surface

r ≈ R
−−−
GM
⇒ Ve = √
R

Now escape velocity at earth surface

−−−−
2GM –
Ve = √ = √2V0
R

14. The work obtained in bringing a body from infinity to a point in gravitationa field is called gravitational potential energy.
Force of attraction between the earth and the object when an object is at the distance a from the center of the earth.
GmM
F =
2
x

Consider the dW is the small amount of work done in bringing the body without acceleration through the very small distance dx.
dW = Fdx
And total work done to bring the body from infinity to point p which is at distance r from the center of the earth.
r
GmM GMm
w = ∫ dx = −
2 r
x
∞

Hence, the gravitational potential energy = − GMm

15. Weight of the body at the earth’s surface

w = mg = 250 N ...(i)
Acceleration due to gravity at depth d from the earth’s surface
′ d
g = g (1 − )
R

here, d = R

2
R/2
∴ g
′
= g (1 −
R
) = g (1 − 1

2
)

g
′
⇒ g =
2

∴ The weight of the body at depth d

mg
′ ′
⇒ w = mg =
2

Using Eq. (i) we get

4 / 13
 w =
′ 250

2
= 125 N
∴ Weight of the body will be 125 N.
16. a. T = 7h 39min = 459 × 60s , orbital radius is given by R and
R = 9.4 × 10 km = 9.4 × 10 m
3 6
,M = ? m

∴ Mass of mars is given by, M m = 4π

2

⋅
R
3

G T 2

2 6 3
4×(3.14) × (9.4× 10 )

= −11 2
6.67× 10 ×(459×60)

= 6.48 × 10 23
kg
2 3
Tm R

b. Using Kepler's third law, =

MS

2 3
T R
ES

where RMS is the mass-sun distance RES is the earth-sun distance

3/2
∴ Tm = (1.52) × 365 = 684 days

17.

i. As AO = BO = CO = 1 m, hence we have
∣−→∣ ∣−→∣ ∣−→∣
GM⋅2M 2
∣FA ∣ = ∣FB ∣ = ∣FC ∣ = = 2GM
2
∣ ∣ ∣ ∣ ∣ ∣ (1)

If we consider direction parallel to BC as x-axis and perpendicular direction as y-axis, then as shown in figure, we have
⃗ 2^
F A = 2GM j

⃗ 2 ∘^ 2 ∘^
F B = (−2GM cos 30 i − 2GM sin 30 j )

−→
and F C = (2GM
2 ∘^
cos 30 i − 2GM
2 ∘^
sin 30 j )

Therefore, the net force on mass 2M placed at the centroid O is given by,
⃗ ⃗ ⃗ ⃗
F = FA + FB + FC

√3 √3
2 ^ 1 ^ ^ 1 ^
= 2GM [j + (− i − i) + ( i − j )]
2 2 2 2

= 0

ii. If mass at the centroid of the triangle gets doubled, even then the net force on it will be zero.
18. Orbital period of I0 is given by, Tl0 = 1.769days = 1.769 × 24 × 60 × 60s
Orbital radius of I , R = 4.22 × 10 m 0 lo
8

Satellite I0 is revolving around the Jupiter

Mass of the latter is given by the relation:
2 3
4π R

MJ =
±

2
…..(i)
GT
fo

Where,
MJ = is the Mass of Jupiter
G = is the Universal gravitational constant
The orbital period of the Earth is given by ,
T = 365.25 days = = 365.25 × 25 × 60 × 60s
θ

Orbital radius of the Earth,

11
Rθ = 1AU = 1.496 × 10 m

Mass of sun is given as:

2 3
4π Re
Ms =
2
…(ii)
GT
θ
2 3 3 3 2
Mx 4π R GT R T
θ lo θ lo
∴ = × = ×
MJ 2 2 3 3 2
GT 4π R R T
θ lo lo θ

5 / 13
 2 3
11
1.769×24×60×60 1.496×10
= ( ) × ( )
365.25×24×60×60 8
4.22×10

= 1045.04
Hence the ratio of mass to the sun is given by:
Ms
∴ ∼ 1000
MJ

Ms ∼ 1000 × MJ
Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.
19. Satellite that appears to be located at a fixed point in space when viewed from the earth's surface.
r
1
U = GMm ∫ dx
2
x
∞
r
1
U = GMm∣
∣−
∣
∣
x ∞
1 1
U = GMm [− + ]
r ∞
GMm
U= −
r

Kinetic energy KE = 1

2
mv
2

−−−
But v = √ GM

1 GM
K. E = 2
m(
r
)

GMm
KE =
2r

Total energy of satellite E = U + v

GMm GMm
E= − +
r 2r
GMm
E= −
2r

T.E. < 0 or negative, this means the satellite is bound to the earth through gravity.

2
mv
Fc =
r
2
GMm mv
=
2 r
r
−−−
GM
v = √
r
−−−
−−−−−−−
Angular momentum, L = mvr = m√ GM

r
× r = √GM m r
2

−−−−−−
2
L = √GMm × √r or L ∝ √r
20. Third law of Kepler. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its
orbit. This captures the relationship between the distance of planets from the Sun, and their orbital periods.
rp = 2R, ra = 6R
Hence, rp = a (1 - e) = 2R …(i)
ra = a (1 + e) = 6R …(ii)
On dividing (i) by (ii)
1−e 2
=
1+e 6

3-3e=1+e
1
4e = 2 ⇒ e =
2

There is not external force or torque on system.

So by the law of conservation of angular momentum.
L1 = L2
ma va ra = mp vp rp , ma = mp = m = mass of satellite
va rp
2R 1
∴ = = =
vp ra 6R 3

So, v = 3vp a

Apply conservation of energy at apogee and perigee …(iii)

1 2 GMm 1 2 GMm
mvp − = mva −
2 rp 2 ra

6 / 13
 Multiplying 2

m
to both side and putting rp = 2R and ra = 6R
2
vp −
2GM

2R
= va −
2 2GM

6R
(where M is mass of earth)
vp
va =
3

2 2 GM 1 GM
∴ vp − va = −
R 3 R
2
vp
2 GM 1
vp − ( ) = [1 − ]
3 R 3

2 1 GM 2
vp [1 − ] = ⋅
9 R 3

2 8 GM 2
vp = ⋅
9 R 3

2 GM 2 9 3 GM
vp = × =
R 3 8 4 R
−−−−−−−−−−−−−−
−−−−− −11 24
3 GM 3×6.67× 10 ×6× 10
vp = √ = √
4 R 6
4×6.6×10

−−−−−−−−−−−−
24−6−11−1
9×667×10
= √
128
−−−−−−−−−− −−−−−−−− −
18−11−1
6003×10 6
vp = √ = √46.89 × 10
128

= 6.85 × 103 m/s = 6.85 km/s

vp
6.85
va = = = 2.28km/s
3 3
−−− −−−−−−−−−−−
−11
−
24
GM 6.67× 10 ×6× 10
vc = √ = √
r 6R
−−−−−−−−−−
24−11
−−−−−−−−−−
6.67×6×10 667 13−5
= √ = √ × 10
6 640
6×6.4×10

−−−−−−−−−−−− − −−−−−−−− −
6 6
= √1.042 × 10 × 10 = √10.42 × 10

vc = 3.23 km/s
Hence to transfer to a circular orbit at apogee we have to boost the velocity by
v0 - va = (3.23 - 2.28) = 0.95 km/s.
21. Mass of each star, M = 2 × 10 30
kg

Radius of each star, r = 10 m 7

Initial potential energy of the stars when they are 10 12

m apart = - GM×M

12
=- GM

12
10 10

[distance between two stars = 10 12

m ]
When the stars are just going to collide, the distance between their centres = twice the radius of each star = 2r = 2 × 10 7
m
2

Final potential energy of the stars when they are about to collide = −G M×M

7
=- GM

7
2×10 2×10

Change in potential energy of stars

2 2 2 2

=- GM

12
- (− GM

7
) = GM

7
- GM

12
10 2×10 2×10 10
2 2 2
GM GM GM
≈ [as << ]
7 12 7
2×10 10 2×10

Suppose v be the speed of each star just before colliding,

Final KE of the stars = 2 × M v = M v 1

2
2 2

Initial KE of the stars= 0

(when the stars are initially 10 m apart, their speeds are negligible). 12

Change in KE of the stars = M v 2

Using the law of conservation of energy, from Eqs.(i) and (ii),

2
GM 2
⇒ = Mv
7
2×10
−−−−−
GM
v = √
7
2×10
−−−−−−−−−−−−−−
−11 30
6.67× 10 ×(2× 10 )

v = √
7
2×10

6
v = 2.6 × 10 m/s

22. Consider a small element of the ring of mass dM at point A. Distance between dM and m is x.

Also x 2
= r
2
+ h
2

G(dM)m
Gravitational force between dM and m, dF =
x2

7 / 13
 dF has two components dF cos θ along PO and dF sin θ perpendicular to PO.
Due to the symmetry of the ring, ∫ dF sin θ = 0
G(dM)m
Net force on mass m due to ring is given by F = ∫ dF cos θ = ∫
2
⋅
h

x
x

=
Gmh
∫ dM =
GMmh
=
GMmh
.......(1)
x3 x3 3/2
2
( r2 + h )

When mass is displaced upto distance 2h then;

F =
′
F =
GMm2h
..........(2)
3/2
′ GMm2h

3/2
2 2
( r2 +(2h ) ) ( r2 +4h )

When h = r from eq. (1)

F =
GMmr

3/2
=
GMm

2
......(3)
2 2 2√2r
(r + r )

from eq.(2)
GMmr 2GMm
F
′
=
3/2
=
2
....... (4)
2 2 5√5r
(r +4r )

Dividing eq(4) and eq(3) we have

′ 4√2
F
=
F 5√5

4√2
or F ′
= F , is the gravitational force between m and ring at distance 2h.
5√5

23. When an object falls freely towards the surface of earth from a certain height, then its velocity changes and this change in velocity
produces acceleration in the object which is known as acceleration due to gravity denoted by g.
g value is less than at the pole as compared to the equator. Variation of g due to Rotation of Earth: It decreases as the rotation of
the Earth increases.
Consider a body of mass 'm' at point P.
w.k.t.,
Fne t
g =
m
2 2
mg−mω R cos λ
gλ =
m

2 2
gλ = g − ω R cos λ

Case (i):
At equator, λ = 0
2
gλ = g − ω R

If ′
ω
′
of earth increases there is a possibility of 'g' at equator becoming zero.
i.e., 0 = g - ω R ⇒ g = ω R 2 2

−
−
g
ω = √
R

−
−
g
i.e., when ω = √ R
, g at equator will become zero.
Case (ii):
At poles, λ = 90o, ∴ g = 0 λ

i.e., 'g' at poles is independent of rotation of earth.

24. Let the gravitational field due to the two stars be zero at some point O lying at a distance x from the centre of smaller star.

(16M)m
Then G Mm

2
= G
2
x (10a − x)

1 16
or 2
=
2
x (10a − x)

or 16x2 = (10a - x)2

or 4x = ± (10a - x)
The negative sign is inadmissible, so x = 2 a
The body of mass m, when fired from point P lying on the surface of heavier star, must cross the threshold (the point O),
otherwise, it would return back.
The gravitational potential energies when the body of mass m lies at positions P and O are given by
GMm G × 16M × m 65GMm
Up = − − = −
8a 2a 8a

8 / 13
 GMm G × 16M × m 5GMm
UO = − − = −
2a 8a 2a

∴ Increase in potential energy,

5GMm 65GMm 45GMm
ΔU = Up - UO = − 2a
+
8a
=
8a

If v is the minimum speed with which the body is fired from P so as to reach O, then
1 2 45GMm
mv = ΔU =
2 8a
−−−−− −−−−
45 GM 3 5GM
or v = √ 4 a
=
2
√
a

25. Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus.
Let a planet be moving in an elliptical orbit around the sun. Since the mass of the sun is very large than that of the planet, the sun
will exert a gravitational force (attractive) on the planet.

GmMs
F =
2
...(1)
R

Since the eccentricities are small and sun is more massive than the planets, the path can be assumed as approximately a circle. The
centripetal force, exerted by the sun on the planet, is given by

2 2

Fc =
mv

R
=
m

R
(
2πR

T
) ...(2)
From (1) & (2)
2 2
GmM m 4π R
= ( )
2 R 2
R
T

2 2
T 4π
=
3 GM
R

Here, 4π , and M are constants, therefore,

2

2
T

3
= constant.
R

9 / 13
 T2 = (constant) R3
T2 ∝ R3
This is the law of periods.

Areal velocity
2 2
πR πR
= = = constant.
T 3
4π 2 R
√( )
GM

26. In Figure, three mass points, each of mass m, are placed at three vertices of an equilateral △ ABC of side l. If O is the centroid of
the triangle, then
OA = OB = OC.

From the right △ODB,

l

∘ BD 2
cos 30 = =
OB OB
1 l

or OB = cos 30
2

∘
=
√3
2
=
l

√3

Gravitational fields at O due to mass points at A, B and C are as follows:

→
E1 = Gm

2
=
Gm

l 2
=
3Gm

2
, along OA
(OA) ( ) l
√3

→
Gm 3Gm
E2 = 2
=
2
, along OB
(OB) l

→
E3 = Gm

2
=
3Gm

2
, along OC
(OC) l

Angle between E⃗ and E⃗ is 120°. Their Resultant is

1 2
−−−−−−−−−−−−−−−−−−−−−
E = √E 2
1
+ E
2
2
+ 2E1 E2 cos 120
∘

−−−−−−−−−−−−−− →
3Gm 3Gm
= 2
√1 + 1 + 2 × (−
1

2
)=
2
, along OF
l l

Clearly, E is equal and opposite to E⃗ , hence the resultant gravitational field at O is zero.
⃗
3

As the gravitational potential is a scalar quantity, so the total gravitational potential at O is

V = V1 + V2 + V3 = − − −
Gm

OA
Gm

OB
Gm

OC

=− 3Gm

OA
= −
3Gm

l
[∵ OA = OB = OC =
l
]
√3

√3

– Gm
or V = −3√3 l

27. Mass of each star, M = 2 × 1030 kg

Radius of each star, R = 104 km = 107 m
Distance between the stars, r = 109 km = 1012 m
For negligible speeds, v = 0
total energy of two stars separated at distance r
−GMM 1 2
= + mv
r 2
−GMM
=
r
+ 0 ….(i)
Consider the case when the stars are about to collide:
Velocity of the stars = v|
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = M v + M v 1

2
2 1

2
2 2
= Mv

−GMM
Total potential energy of both stars = 2R

10 / 13
 Total energy of the two stars = M v 2
−
GMM

2R
…..(ii)
Using the law of conservation of energy,
2 GMM −GMA
⇒ Mv − =
2R R

2 −GM GM 1 1
v = + = GM (− + )
r 2R r 2R

= 6.67 × 10-10 × 2 × 10-10 [− 1

12
+
1

7
]
10 2 × 10

= 13.34 × 1019[-10-12 + 5 × 10-8]

19 −12 −8
= 13.34 × 10 [− 10 + 5 × 10 ]

= 13.34 × 1019 × 5 × 10-8 = 6.67 × 1012

−−−−−−−−−
v = √6.67 × 10 = 2.58 × 106 m/s 12

28. Initial velocity of the rocket is given by, v = 2 km/s = 2 × 10 m/s 3

Mass of Mars is given by, M = 6.4 × 10 kg 23

Radius of Mars is given by, R = 3395 km = 3.395 × 10 m 6

−11 2 −2
G = 6.67 × 10 Nm kg

Mass of the rocket is given by = m

K. Ei = mv 1

2
2

−GMm
P. Ei= R

Total initial energy is = ( 1

2
mv
2
−
GMm

R
)

If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in
reaching a height.
Total initial energy available is = [ 80

100
][
1

2
2
mv ] − [
GMm

R
2
] = 0.4mv −
GMm

At the height h, the velocity and hence, the kinetic energy of the rocket will become zero.
−GMm
Total energy of the rocket at height h is = R+h

Applying the law of conservation of energy

2 GMm −GMm
0.4mv − =
R (R+h)

2 GM GM
0.4v = −
R R+h

1 1
= GM ( − )
R R+h

R+h−R
= GM ( )
R(R+h)

GMh
=
R(R+h)

R+h GM
=
h 2
0.4v R
R GM
+ 1 =
h 2
0.4v R
R GM
= − 1
h 2
0.4v R
R R
=
h GM
−1
0.4v2 R
2 2
0.4R v
=
2
GM−0.4v R

6 2 3 2
0.4× (3.395× 10 ) × (2× 10 )
=
−11 23 3 2 6
6.67× 10 ×6.4× 10 −0.4× (2× 10 ) ×(3.355× 10 )

18
18.442×10
=
12 12
42.688× 10 −53432× 10

3
= 495 × 10 m = 495km

GM GM
29. i. g = 2
=
2
(R+h) 2 h
R (1+ )
R

−2
GM h
g = (1 + )
2 R
R
GM
Now, 2
= g0 (surface value)
R
−2
h
∴ g = g0 (1 +
R
) [which < < R]
2h
g ≈ g0 (1 − )
R

[Using binomial theorem (1 + x)n ≈ 1 + nx when |x| < < 1.]

If we go inside the earth at depth h

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 ii. As depth of mine is < < R
Hence we can use approximation formula
Let height be h
We know that for variation of g with height,
2h
g = g0 (1 −
R
) ...(i)
And for variation in g with depth
d
g = g0 (1 −
R
) ...(ii)
Where d is the depth.
According to the question
2h d
g0 (1 − ) = g0 (1 − )
R R

2h d
⇒ =
R R
d 80
⇒ h = = = 40 km
2 2

30. Discovery of Newton's law of gravitation: One day in the year 1665, seeing an apple falling from a tree, Newton was inspired to
think about the law of gravitation. He thought that the force which attracts the apple toward the earth might be the same as the
force attracting the moon toward the earth. By comparing the acceleration due to gravity on the earth with the acceleration
required to keep the moon in orbit around the earth, Newton was able to deduce the law of gravitation as discussed below.
Newton assumed that the moon revolved around the earth in a circular orbit of radius R (= 3.84 × 108m), as shown in Figure.

Period of moon around the earth, T = 27.3 days = 27.3 × 86,400 s

Circumference of orbit
Speed of the moon v =
Orbital period
8

= 1.02 × 103 ms-1

2π ×(3.84 × 10 m)
=
27.3 × 86400 s

The centripetal acceleration of the moon,

3 2
2 (1.02 × 10 )
v
ac = =
R 8
3.84 × 10

= 2.72 × 10-3 ms-2

31. Acceleration due to gravity is the acceleration gained by an object due to gravitational force. Its SI unit is m/s2.
The force (F) of gravitational attraction on a body of mass m due to earth of mass M and radius R is given by
F = G ...(i)
mM

2
R

We know from Newton's second law of motion that the force is the product of mass and acceleration.
∴ F = ma

But the acceleration due to gravity is represented by the symbol g. Therefore, we can write
F = mg ...(ii)
From the equation (i) and (ii), we get
mM GM
mg = G
2
or g = 2
...(iii)
R R

When body is at a distance 'r' from centre of the earth then g = GM

2
.
r

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 The acceleration due to gravity on earth surface is given as
GMe
ge =
2
Re

6 2
2 9.8×(6.37× 10 )
g R
or M e =
s

G
e
=
−11
6.67×10

=6× kg 1024
32. There is a point called the La-Grange L1 point roughly in a million miles from earth towards the sun. It could be considered that
the net gravitational force at this point is zero. Practically it can never be zero. However, there are places where there is so little
gravity that it becomes an insignificant factor, and we consider gravity at that point to be zero.
Given parameters are:
Mass of the Sun, M = 2 × 10 kg s
30

Mass of the Earth, M = 6 × 10 kg θ

24

Orbital radius, r = 1.5 × 10 11

m

Mass of the rocket = m

Let's suppose that x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton's law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the
Earth as:
GmMs GmM
θ
=
2 2
(r−x) x

r−x 2 Ms
( ) =
x M
θ
1
30
r−x 2×10 2
= ( ) = 577.35
x 24
6×10

1.5 × 1011 - x = 577.35 x

577.35 x = 1.5 × 1011
11

= 2.59 × 108 m.
1.5 × 10
x =
578.35

Now we can say that this is the distance at which the gravitational force on the rocket from earth's centre is zero.

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