CH 07
X-PHYSICS
FAHAD AKHTER
CIRCULAR MOTION AND GRAVITATION
UNIFORM CIRCULAR MOTION:
“If a body moves along a circular path with a uniform speed then it is said to be in uniform circular
motion.”
CENTRIPETAL ACCELERATION:
“Acceleration of a body in a circular path is called centripetal acceleration.”
𝐂𝐞𝐧𝐭𝐫𝐢𝐩𝐞𝐭𝐚𝐥 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 =
𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝟐
𝐑𝐚𝐝𝐢𝐮𝐬
𝐚𝐜 =
*Centripetal acceleration represented by ac.
*Unit of centripetal acceleration is m/s2.
𝐕𝟐
𝐫
CENTRIPETAL FORCE:
“A center seeking force which keeps the object moving in a circle is called centripetal force.”
Fc = mac
𝐅𝐜 =
*Centripetal force represented by Fc.
*Unit of centripetal Force is Newton (N).
𝐦𝐕 𝟐
𝐫
Example:
*Electron revolves around the nucleus of atom in circular orbit.
*Planets revolve around the sun.
CENTRIFUGAL FORCE:
“The outward reacting force on the result of centripetal force is called centrifugal force.”
GRAVITATION:
“The force with which earth attracts bodies toward its center is called gravitation.”
NEWTON’S LAW OF UNIVERSAL GRAVITATION:
Statement:
“Everybody in the universe attract every other body with a force which is directly proportional to the
product of their masses and inversely proportional to the square of the distance between their centers.”
Explanation:
Let “m1” and “m2” be the masses of two spherical bodies, such
that their center are separated by a distance “r” , then
according to the Newton’s law of
Universal gravitation the force “F” with which they attract each
other is given by:
F α m1 m2~~~~~~~~eq (i
𝐅𝛂
𝟏
𝐫𝟐
~~~~~~~~eq (ii
By combining eq (i & eq (ii we get:
𝐅𝛂
𝐦𝟏 𝐦𝟐
𝐫𝟐
𝐅 = 𝐆
𝐦𝟏 𝐦𝟐
𝐫𝟐
*Where “G” is the constant of universal gravitational constant and its value is 6.67 x 10-11Nm2/Kg2.
X-PHYSICS
FAHAD AKHTER
GRAVITATIONAL CONSTANT “G”:
The value of G was determined experimentally by Henry Cavendish in 1798 with an instrument called
Cavendish balance.
The Cavendish balance consist of two small spheres of mass
“m” mounted at opposite end of a light horizontal rod of
length “d” which is suspended at its center by a fine vertical
fiber. When two large spheres of mass “M” are brought near
to small sphere at distance “r”, as shown in figure. The force
of gravitational attraction between the spheres produces
torque, which twists the fiber. The twist in fiber is proportional
to the magnitude of torque.
If "
𝑮𝒎𝑴
𝐚𝐧𝐝
𝒓𝟐
" is the gravitational force between the spheres
𝑮𝒎𝑴
x d is the torque which produce twist “θ” the
𝒓𝟐
fiber.
Then
𝑮𝒎𝑴
× 𝒅𝜶𝜽
𝒓𝟐
𝑮𝒎𝑴
× 𝒅 = 𝑪𝜽
𝒓𝟐
Where ‘C’ is a constant of proportionality. Its value depends upon the nature of the material of fiber.
By substituting the value of all known quantities in the above equation the value of “G” can be
calculated.
𝒓𝟐
𝑮 =
𝑪𝜽
𝑴𝒎𝒅
The accepted value of G = 6.67 x 10-11 Nm2/kg2.
MASS OF EARTH:
Consider a body of mass “m” lying on the surface of earth so that the distance between the center of
the earth and the center of the body is equal to the radius “R” of the earth.
The force with which earth pulls the body toward its center is known
as its weight “W” given by:
W=mg
But according to the Newton’s law of gravitation the force “F” with
which earth pulls the body towards its center given by:
𝐅 = 𝐆
𝐦 𝐌𝐞
𝐑𝟐
Where “Me” is the mass of the earth and “F” & “W” represents the same force.
F=W
𝐆
𝐦 𝐌𝐞
𝐦𝐠
𝐑𝟐
𝐆
Where,
G = 6.67 x 10-11 Nm2/Kg2
g = 9.8 m/s2
𝐌𝐞
𝐠
𝐑𝟐
𝐌𝐞 =
𝐠 𝐑𝟐
𝐆
X-PHYSICS
-
FAHAD AKHTER
R = 6.4 x 106 m
Substitute the value of G, g & R we get:
𝐌𝐞 =
𝟗. 𝟖 𝐱 (𝟔. 𝟒 𝐱 𝟏𝟎𝟔 )𝟐
𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏
Me = 6 x 1024 KG
1. Value of ‘g’ decreases with altitude or variation of ‘g’:
Value of g decreases with altitude:
The acceleration of gravity “g” is the acceleration imparted to a body by its own
The acceleration of gravity “g” is the acceleration imparted to a body by its
Weight. Consider a body of mass “m” on the earth's surface. The gravitational force
own weight. Consider a body of mass “m” on the earth's surface. The
Of attraction between the body and earth is equal to the weight of body.
gravitational force of attraction between the body and earth is equal to the
weight of body.
𝑭=𝑾=
𝒎𝒈=
𝒈=
𝑮 𝒎 𝑴𝒆
𝑹𝟐
𝑮 𝒎 𝑴𝒆
𝑹𝟐
𝑮 𝑴𝒆
𝑹𝟐
At any height “h” above the surface of the
earth, “g” is given by:
𝒈𝒉 =
𝑮 𝑴𝒆
(𝑹 + 𝒉)𝟐
Dividing g by gh we get
𝑮 𝑴𝒆
𝒈𝒉 (𝑹 + 𝒉)𝟐
=
𝑮 𝑴𝒆
𝒈
𝑹𝟐
𝒈𝒉
𝑮𝑴𝒆
𝑹𝟐
=
𝑿
𝟐
𝒈
(𝑹 + 𝒉)
𝑮 𝑴𝒆
𝒈𝒉
𝑹𝟐
=
𝒈
(𝑹 + 𝒉)𝟐
𝟏
𝒈𝒉 (𝑹 + 𝒉)𝟐
=
𝒈
𝑹𝟐
𝟏
𝒈𝒉 𝑹𝟐 + 𝟐𝑹 𝒉 + 𝒉
=
𝒈
𝑹𝟐
Since, "h" is very small as Compared to Re,
therefore,
h2
Re 2
is negligible.
𝒈𝒉
𝟏
𝒈 = (𝟏 + 𝟐 𝒉)
𝑹
𝒈𝒉
𝟐𝒉 −𝟏
= (𝟏 + )
𝒈
𝑹
But by binomial expansion,
The above result is obtained by neglecting the
terms with higher power of “Re”.
Therefore,
𝒈𝒉
𝟐𝒉
= (𝟏 −
)
𝒈
𝑹
𝒈𝒉 = 𝒈
(𝟏 − 𝟐𝒉)
𝑹
𝐠𝐡 < g
Hence the equation shows that value of g
decreases with height.
𝒈𝒉
𝟏
= 𝟐
𝒈
𝑹
𝟐 𝑹𝟐 𝒉 𝒉𝟐
+
+ 𝟐
𝑹𝟐
𝑹𝟐
𝑹
𝒈𝒉
=
𝒈
𝟏
𝟐 𝒉 𝒉𝟐
𝟏+
+
𝟐 𝑹 𝑹𝟐
2. VALUE OF ‘g’ DECREASES WITH ALTITUDE OR VARIATION OF ‘g’:
The acceleration of gravity “g” is the acceleration imparted to a body by its own Weight. Consider a
body of mass “m” on the earth's surface. The gravitational force Of attraction between the body and
earth is equal to the weight of body.
X-PHYSICS
FAHAD AKHTER
Weight of body.
𝑭=𝑾=
𝒎𝒈=
𝒈=
Where G and Me are constant,
𝑮 𝒎 𝑴𝒆
𝑹𝟐
𝑮 𝒎 𝑴𝒆
𝑹𝟐
𝑮 𝑴𝒆
𝑹𝟐
𝒈 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝟏
𝑹𝟐
𝟏
𝑹𝟐
Above relation shows that value ‘g’ decreases with altitude.
𝒈∝
ARTIFICIAL SATELLITE:
“The objects which are used for communication and space research, revolve around the earth under
the force of gravity, which provide the necessary centripetal force, made by human not nature is
called artificial satellite.”
ORBITAL VELOCITY:
Consider a satellite of mass “m” moving in an orbit of radius “r” with velocity “V” around the earth. The
gravitational force of attraction between the satellite and the earth provide necessary centripetal force.
If Me is the mass of the earth, than
𝑮𝑴𝒆 𝒎
𝑮𝒓𝒂𝒗𝒊𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒇𝒐𝒓𝒄𝒆 =
𝒂𝒏𝒅
𝒓𝟐
𝑪𝒆𝒏𝒕𝒓𝒊𝒑𝒆𝒕𝒂𝒍 𝒇𝒐𝒓𝒄𝒆 =
𝒎𝑽𝟐
𝒓
𝑮𝑴𝒆 𝒎
𝒎𝑽𝟐
=
𝟐
𝒓
𝒓
Or
𝑽𝟐 =
𝑮𝑴𝒆
− − − − − − − − − (𝒊)
𝒓
If ‘g’ is the acceleration due to gravity at the position of satellite then
𝒎𝒈 =
𝒈𝒓 =
𝑮𝑴𝒆 𝒎
𝒓𝟐
𝑮𝑴𝒆
− − − − − − − − − (𝒊𝒊)
𝒓
Comparing equation (i) and (ii) we have
𝐕 𝟐 = 𝐠𝐫 𝐨𝐫 𝐕 = √𝐠𝐫
If satellite revolving near to the surface of the earth then r = Re = radius of earth.
CH 07
∴ 𝑽 = √𝒈𝑹𝒆
MULTIPLE CHOICE QUESTIONS (MCQ’S)
V = 7920 km/s
1.
The radius of earth is equal to
i) 6.4 x 106m
ii) 6 x 1024m
2. The mass of earth is equal to
i) 6 x 1024Kg
ii) 6 x 109kg
iii) 6.9 x 1027m
iii) 6 x 1027kg
X-PHYSICS
FAHAD AKHTER
3. The Value of gravitational constant is equal to
i) 6.67 x 10-11N-m2/Kg2
ii) 7.67 x 10-11N-m2/Kg2
4.
iii) 8.67 x 10-11N-m2/Kg2
If stone is tied to the end of the string and whirled in a circle the tension in the string provides
i) Centripetal force
ii) Pressure
5. G is called
.
i) Acceleration due to gravity
iii) Normal reaction
ii) Gravitational force
iii) Gravitational constant
6.
The gravitational force between two bodies depends upon the product of their mass, in addition to
them
i) Distance between
ii) Shape of bodies
iii) Sum of mass
7. Centripetal force =
i) mr/v2
ii) rv2/m
8. The equation ME =
i) gG/RE2
iii) mv2/r
ii) GRE2/g
is used to determined the mass of earth.
iii) gRE2/G
9. The unit of centripetal force is
i) Joule
ii) Watt
iii) Newton
10. g is called
i) Acceleration due to gravity
CH 07
ii) Gravitational force
iii) Gravitational constant
PAST TEN YEARS QUESTIONS

(2003)
Q: Define the following I) Centripetal force.
Q: State the following laws I) Newton’s law of universal gravitation.

(2004)
Q: Define the following I) Centripetal force.
(BLANK): The magnitude of centripetal acceleration ac = V2/r.
Q: Derive the equation Me = gR2/G to determine the mass of earth by applying law of universal
gravitation.

(2005)
Q: State Newton’s law of universal gravitation and drive F = Gm1m2/r2.
Q: Calculate the tension in a string 50m long when it is tied at its end to mass of 0.2 Kg & whirled in a
circle with a speed of 100 m/s.

(2006)
Q: What is the mass of an object whose weight is 294 N at the surface of the earth?
Q: State the Newton’s law of universal gravitation and derive Me = gR2/G.

(2007)
(BLANK) The value of G is 6.67 x 10-11 Nm2/Kg2.
Q: Define centripetal force, centripetal acceleration and write their formula and unit?

(2008)
Q: A mass of 10 kg at the end of the string is being whirled in a circle of radius 5 cm with a speed of
4 m/s. what will be centripetal force.
Q: Write two points of difference between the following: I) G & g.
(BLANK): The point at which whole weight of the body appears to acts is called Center of gravity.
Q: Define Newton’s law of universal of gravitation and find the equation for the mass of the earth
with the help of F = Gm1m2/r2.

(2009)
X-PHYSICS
FAHAD AKHTER
Q: State Newton’s law of universal gravitation and derive F = Gm1m2/r2.
(BLANK): The value of G is 6.67 x 10-11 Nm2/Kg2.
Q: Define centripetal force. What are the factors on which it depends? Write down the formula of
centripetal force.
Q: Define: I) Center of gravity II) Torque.


(2010)
Q: A stone of 200 gm mass tied to one end of a string of length 50 cm is being whirled from the
other end in a circle with the constant speed of 2m/s. Find the tension in the string.
Q: A force of 50 N acts on a body. If the moment arm is 0.5 m. fin the value of torque.
(2011 & 2012)(No Question In This Year)
Easy to set rules but difficult to follow them.