Circular Motion and Planets

Angles

Can be measured in degrees or radians.
Angle in radians = arc length / radius length
? = s / r
from degrees to radians ~> 360 = 2 ? rad

Angular Velocity

? = ? / t
(? = angular velocity; ? = change in angular displacement; t = time taken for change)
Vector quantity, written in rad/s.

Derivation of relationship between angular and linear velocity

(v = linear velocity; ? = angular velocity; r = radius)

Periodic Time "T

The periodic time (or more simply the period) is the time taken to complete one full circular motion
by definition ? = ? / t therefore for a full circular motion (? = 2 ? rad) ? = 2 ? / T
if we rearrange we get T = 2 ? / ?

A particle travelling in curcular motion undergoes an angular displacement of 180� in a time of 5 seconds. Calculate the angular velocity

Change the degrees to radians 360� = 2 ?
Dividing by 2 gives 180� = ?
By definition ? = ? / t
? = ? / 5
? = 0.628 rad/s

Calculate the angular velocity of a particle travelling at 7 m/s in a circle of radius 20 m

? = v / r
? = 7 / 20
? = 0.35 rad/s

A disk of diameter 30 cm rotates at 33.3 r.p.m. Calculate
(i) the angular velocity in rad/s
(ii) the linear speed of a point on the edge of the disk

(i) 33.3 r.p.m. = 33.3 revolutions / minute
= 33.3 x 2? rad / 60 sec
= 3.49 rad/s
(ii) recall v = r?
v = 0.15 x 3.49
v = 0.52 m/s

(i) Calculate the angular velocity on the earth's rotation
(ii) If the radius of the earth at the equator is 6.4 x 10^6 m calculate the linear velocity of a point on the surface of the earth at the equator

(i) recall ? = 2 ? / T
? = 2 ? / 24 x 60 x 60
? = 7.27 x 10^-5 rad/s
(ii) recall v = r?
v = (6.4 x 10^6) x (7.27 x 10^-5 )
v = 465 m/s

Acceleration for uniform circular motion

When we study circular motion of moving objects, they don't speed up or slow down. How do they accererate then?
They continually change direction in a process called centripetal acceleration
a = v^2 / r or a = r x ?^2
(a = acceleration; r = radius; v = li

Centripetal acceleration

The acceleration acting towards the centre of a circle for an object undergoing uniform circular motion

Centripetal force

The force acting towards the centre if the circle for an object undergoing circular motion
F = m x v^2 / r & F = m x r x ?^2
(F = centripetal force; m = mass; v = linear velocity; r = radius; ? = angular velocity)

A car of mass 1600 kg travels in circular motion at a uniform velocity of 24 m/s. The diameter of the circular motion is 30 m. Calculate the centripetal force needed to maintain this circular motion.

Diameter to radius ~> 30 / 2 = 15 m
Recall F = m x v^2 / r
F = 1600 x (24)^2 / 15
F = 61, 440 N

A stone of mass 5 kg is tied to the end of a string and it undergoes circular motion in a horizontal plane. The radius on the circular motion is 1.6 m. The angular speed of the stone and string is 6 rad/s. Calculate the tension in the string.

The tension is the force that maintains the string, i.e. the centripetal force.
Therefore we use T = m x r x ?^2
T = 5 x 1.6 x (6)^2
T = 288 N

A stone of mass 5 kg is tied to the end of a string and it undergoes circular motion in a vertical plane. The radius on the circular motion is 1.6 m. The angular speed of the stone and string is 6 rad/s. Calculate the tension in the string at the
(i) top

(i) T + mg = m x v^2 / r
or
T + mg = m x r x ?^2
Since we have the angular velocity we use:
T + mg = m x r x ?^2
T + 5(9.8) = 5 x 1.6 x (6)^2
T + 49 = 288
T = 239 N Minimum Tension at the top
(ii) T - mg = m x v^2 / r
or
T - mg = m x r x ?^2
Since we have

Newton's law of universal gravitation

Any two point masses in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F = G x m1 x m2 / d^2
(m1 m2 = mass values; d = distan

Relationship between g and G

When the force of gravity between an object of mass "m" and a planet of mass "M" is examined, we use "R" for the radius of the planet and "h" for the distance between the surface of the planet and the object. This distance is called the height.

Relationship (formula) between g and G
(i) on the planet surface
(ii) above the planet surface

(i) m x g = G x M x m / R^2
g = G x M / R^2
(ii) m x g = G x M x m / {R + h}^2
g = G x M / {R + h}^2
Note: log books have both as g = G x M / d^2. if d is given for above surface, remember h.

Calculate the acceleration due to gravity
(i) on the surface of the earth
(ii) 200 km above the surface of the earth
(G = 6.7 x 10^-11 Nm^2 kg^-2; radius of earth = 6.4 x 10^6 m; mass of earth = 6 x 10^24 kg)

(i) g = G x M / R^2
g = (6.7 x 10^-11) x (6 x 10^24) / (6.4 x 10^6)^2
g = 9.81 m/s^2
(ii) 200 km = 200 x 10^3 m
g = G x M / {R + h}^2
g = (6.7 x 10^-11) x (6 x 10^24) / (6.4 x 10^6 + 200 x 10^3 )^2
g = 9.23 m/s^2

Calculate the gravitational attraction between
(i) the earth and the moon
(ii) the sun and the moon
(G = 6.7 x 10^-11 Nm^2 kg^-2; mass of earth = 6 x 10^24 kg; mass of moon = 7.4 x 10^22 kg; mass of earth = 2 x 10^30 kg; distance between earth and moon =

(i) 375,000 km = 375,000 x 10^3 m
g = G x M / d^2
g = (6.7 x 10^-11) x (6 x 10^24) x (7.4 x 10^22) / (375,000 x 10^3)^2
g = 2.12 x 10^20 N
(ii) 1.35 x 10^8 km = 1.35 x 10^11 m
g = G x M / d^2
g = (6.7 x 10^-11) x (2 x 10^30) x (7.4 x 10^22) / (1.35 x 10^1

The international space station (ISS) moves in a circular orbit around the equator at a height of 400 km
(i) Calculate the acceleration due to gravity at a point of 400 km above the surface of the earth
(ii) Explain why an astronaut appears weightless in

(i) g = G x M / {R + h}^2
g = (6.7 x 10^-11) x (6 x 10^24) / (6.4 x 10^6 + 400 x 10^3 )^2
g = 8.69 m/s^2
(ii) the value for "g" is not zero, so the astronaut does have a weight. However, they do not feel this weight as the space station is in free fall to

How can the ISS be in a state of free fall towards the earth

The ISS is projected along a "tangent" line, it moves along this line and falls to earth at the same time. The momentum of the ISS moves it along another "tangent" line. The combined motion continues on and on, the distance it falls to earth is just enoug

Satellites

The moon is a satellite of the earth
The earth is satellite of the sun
The ISS is a satellite of the earth
The periodic time squared is proportional to the radius of orbit cubed
periodic time formula ~> T^2 = 4 x ?^2 x R^3 / G x M

Derivation of the periodic time squared is proportional to the radius of orbit cubed

Gravitational force between = centriprtal force
satellite and planet keeping satellite in
orbit around planet
G x M x m / R^2 = m x R x ?^2
G x M / R^2 = R { 2 ? / T}^2
G x M = R^3 x (4 x ?^2 / T^2)
T^2 = 4 x ?^2 x R^3 / G x M
T^2 ? R^3

The period of a satellite that orbits Saturn is 380 hours. Calculate the radius of the satellite's orbit around Saturn
(G = 6.7 x 10^-11 Nm^2 kg^-2; mass of saturn = 5.7 x 10^26 kg)

Rearrange T^2 = 4 x ?^2 x R^3 / G x M
To get R^3 = G x M x T^2 / 4 x ?^2
R^3 = (6.7 x 10^-11) x (5.7 x 10^26) x (380 x 60
x 60)^2 / 4 x ?^2
R^3 = 1.81 x 10^27
R = 1.219 x 10^9 m

The ISS orbits the earth at a height of 400 km. Calculate the period of an orbit of the ISS.
(G = 6.7 x 10^-11 Nm^2 kg^-2; radius of earth = 6.4 x 10^6 m; mass of earth = 6 x 10^24 kg)

T^2 = 4 x ?^2 x R^3 / G x M
T^2 = 4 x ?^2 x (6.4 x 10^6)^3 / (6.7 x 10^-11) x (6 x 10^24)
T^2 = 3.087 x 10^7
T = 5557 s
T = 1.54 hours

Geostationary satellites

Give the illusion of being stationary above a particular point on the earth. To an observer on the earth it seems the satellite is always in the same place. Used for communication
For a satellite to be geostationary relative to the earth it must:
Orbit th

Calculate the height of a geostationary satellite above the surface of the earth.
(G = 6.7 x 10^-11 Nm^2 kg^-2; radius of earth = 6.4 x 10^6 m; mass of earth = 6 x 10^24 kg)

R^3 = G x M x T^2 / 4 x ?^2
R^3 = (6.7 x 10^-11) x (6 x 10^24) x (24 x 60 x 60)^2 / 4 x
?^2
R^3 = 7.6 x 10^22
R^3 = 4.24 x 10^7 m
The height above the earth is
h = radius of orbit - radius of earth
h = (4.24 x 10^7) - (6 x 10^6)
h = 3.6 x 10^7 m

Speed of a satellite

When given the mass of the plaent:
Gravitational force between = Centripetal force keeping
satellite and plenty satellite in orbit around
planet
G x M x m / R^2 = m x v^2 / R
G x M / R = v^2
v = ?(G x M / R)
When given the periodic time of the satellite:

The distance from the earth to the sun is 1.5 x 10^11 m. The mass of the sun is 2 x 10^30 kg. Calculate the speed of the earth as it orbits the sun.
(G = 6.7 x 10^-11 Nm^2 kg^-2)

v=?(G x M / R)
v = ?( 6.7 x 10^-11 x 2 x 10^30 / 1.5 x 10^11)
v = 2.99 x 10^4
v = 3 x 10^4 m/s

Experiment to show the relationship between period and length for a simple pendulum (and hence calculate g)

Allow the pendulum to swing from side to side through a small angle, using a varied length in order to gather a set of data.
Measure and record the length of the pendulum from the point of suspension to the centre of gravity of the bob. Measure and record

Accuracy with the experiment to show the relationship between period and length for a simple pendulum

Measure 40 oscillations rather than just one to avoid large percentage errors
Use larger values for length to ensure lower percentage errors
Avoid the error of parallax when measuring the length of the pendulum with the metre stick
Use a heavy bob to keep

Calculation of g (experiment to show the relationship between period and length for a simple pendulum)

The pendulum experiment may be used to calculate the value of the acceleration due to gravity
T = 2 x? ?l / g
T^2 = 4 x ?^2 x l / g
g = 4 x ?^2 x l / T^2
g = 4 x ?^2 x slope
slope = l / T^2

A student gathered the following data when investigating the relationship between the period and length of a single pendulum. The length of the pendulum was l. The time for 30 oscillations was given by t. By drawing a suitable graph, calculate the acceler

l / m | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 |
T / s | 1. 28 | 1.42 | 1.58 | 1.72 | 1.86 | 1.93 | 2.0 |
T^2 / s^2 | 1. 64 | 2.01 | 2.5 | 2.96 | 3.31 | 3.72 | 4.0 |
slope of the graph = y2 - y1 / x2 - x1
= 0.66 - 0.2 / 2.7 - 0.8
= 0.242
g = 4 x ?^2 x sl