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Newton’s Law Of Universal Gravitation
Newton’s Law Of Universal Gravitation states that every particle in the Universe
attracts every other particle 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
𝑚1 𝑚2
𝑟2
, where:
•
•
•
•
•
F is magnitude of the gravitational force between the two point masses,
G is the gravitational constant,
m1 is the mass of the first point mass,
m2 is the mass of the second point mass,
r is the distance between the two point masses.
The gravitational force is inversely proportional to the square of the separation of the
particles.
F∝ 𝑟12
* Only applicable to point masses
Gravitational field strength
Gravitational field strength at a point is defined as the gravitational force per unit
mass acting on a mass placed at that point.
𝐹
𝑚
𝑀
𝑔=𝐺 2
𝑟
𝑔=
Units of g is Nkg−1 or ms−2.
It is independent of test mass.
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Gravitational Field Lines
Gravitational field lines are always pointing towards the center of mass. Magnitude of
gravitational field is proportional to the number of field lines per unit area through a
surface perpendicular to the lines. Therefore, g is larger when lines are close together
and smaller when they are far apart.
Gravitational field strength close to the Earth can be considered as uniform as the
Earth surface is relatively flat at a point.
Variation of gravitational field strength on Earth
•
•
•
Earth is not a perfect sphere (ellipsoid, flattened at the poles). Gravitational field
strength increases(at sea level) when one moves from equator to poles.
Density of Earth is not uniform.
Earth is rotating about an axis passing through the poles. Gravitational Pull on a
body at equator has to provide for the centripetal acceleration required for
rotation. Hence the measured gravitational acceleration is smaller.
Weightless
Weight of an object is the gravitational force exerted on the object by the Earth’s
gravitational field.
An object is only truly weightless when there is no gravitational force acting on it.
(Only true when object is infinitely far away from other masses where no gravitational
field acts on it)
An object is considered to be apparently weightless when it exerts no force on its
support.
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Gravitational Potential Energy
Gravitational potential energy, U of a point mass m, in a gravitational field, is the
work done by an external force in bringing that point mass from infinity to that point.
𝑈=𝐺
𝑀𝑚
𝑟
Units of U is Joules (J).
Reasons for a negative sign in the equation:
•
•
Zero (reference point) for gravitational potential energy is set to be at infinity.
The gravitational pull is an attractive force. Since gravitational potential is
defined to be work done per unit mass by an external force to bring a mass from
infinity to a point, the external force has to do work against the attractive field.
As external force is opposite to the direction of displacement, values of
gravitational potential are negative.
Gravitational Potential
Gravitational potential, ϕ, at a point in a gravitational field is the work done per unit
mass, by an external force, in bringing the mass from infinity to that point.
ϕ=–G
Units for ϕ is JKg−1JKg−1.
At infinity, ϕ is assumed to be zero.
𝑀
𝑟
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Escape Speed
Escape speed is the minimum speed with which a mass should be projected from the
Earth’s surface in order to escape Earth’s gravitation field.
Vescape=√2gRE
, where
•
•
•
V is escape speed,
g is gravitational field strength,
R is radius of the Earth.
Derivation of Escape Speed From Earth
We know that:
Total Energy at infinity=0
Hence,
Kinetic energy + Potential energy = 0
(fill in)
where v is the velocity of the object
m is the mass of the object
M is the mass of Earth
R is the radius of the Earth
G is the universal gravitation constant
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From Gravitational Field Strength, we know that
𝑀
𝑔 = 𝐺 2. Substitute this into the equation above, we will have:
𝑟
(fill in)
In the context of this derivation, we have:
Vescape=√2gRE
Equations relating to U, g and F
U=mϕ
Fg= ( )
g=( )
(
) is known as potential gradient.
Gravitational field strength points in direction of decreasing ϕ.
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Questions for Gravitation (JC) Set 1
Imagine if the Moon doubles in mass, which of the following is likely to happen?
1.
2.
3.
4.
Each month will be longer than 30 days.
Nights will be a lot brighter since the Moon is nearer to the Earth
Each day will be longer since the period of the Moon increases.
The Moon continues to revolve around the Earth in the same orbit.
T2 = kr3 therefore period and orbital radius is independent of mass.
Answer: 4
Which quantity is not necessarily the same for satellites that are in geostationary
orbits around the Earth?
1.
2.
3.
4.
Angular velocity
Kinetic energy
Centripetal acceleration
Orbital period
Period of geostationary satellite’s orbit is the same as Earth (24 hours) and hence, its
angular velocity is also constant. It can also be found that the radius of the
geostationary orbit is fixed using R3∝T2. Hence centripetal acceleration is also
constant. The only quantity that varies is the kinetic energy because it depends on
the mass of the individual satellites.
Answer: 2
Mars is known to possess two satellites, Phobos and Deimos. The former is at a
distance of 9,500 km from the centre of Mars and the latter at a distance of 24,500
km. Find the ratio of the period of Phobos to that of Deimos in their revolutions
around Mars.
Tphobos
Tdeimos
3
= √T𝑟𝑟 3
phobos
deimos
=√
95003
245003
= 0.24
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Answer: 0.24
Assuming the Earth to be a uniform sphere rotating about an axis through the
poles, the apparent weight of a body at the Equator compared with it’s apparent
weight at the poles would be
1. greater, because the angular velocity of the Earth is greater at the Equator
than at the pole
2. greater, because the weight at the Equator is given by the sum of the
gravitational attraction of the Earth and the centripetal force due to the
circular motion of the body
3. smaller, because the gravitational attraction of the Earth must provide both
the weight and the centripetal force due to the circular motion of the body
4. smaller, because the gravitational attraction at the pole is greater than that at
the Equator
The Earth spins about the axis through which its poles pass. A person at the Equator
undergoes circular motion while a person at the poles does not.
Answer: 3
a) The Moon is orbiting at a certain height above the Earth. The Earth exerts a
force of 1.99 x 1020 N on the Moon.
Given: mass of the Earth = 5.98 x 1024 kg
mass of the Moon = 7.35 x 1022 kg
radius of the Earth = 6.38 x 106 m
i) Find the height above the Earth’s surface where the Moon is orbiting.
ii) Using Newton’s Second Law of motion, calculate the linear velocity of the
Moon.
iii) Calculate the centripetal acceleration of the Moon.
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b) Most artificial satellites are placed in Low Earth Orbits (LEO), at a height
typically around 200 km – 1200 km above the Earth’s surface. A 700 kg satellite
is in the Low Earth Orbit at a height of 1200 km above the Earth’s surface.
i) What is the change in gravitational potential energy of the satellite before it is
launched and when it is in orbit?
ii) Calculate the period of the satellite. Express your answer in hours.
iii) Satellites in LEO are used in telecommunications. State one difference
between the LEO and geostationary satellites, and an advantage of using LEO
satellites in telecommunications.
(a) i) F = G(MEMM/r2)
r = 3.8382 x 108 m
Height above Earth’s surface = 3.8382 x 108 – 6.38 x 106
= 3.77 x 108 m
ii) By Newton’s second law of motion,
Fg = Fc
v = (GME/r)1/2
v = 1.02 x 103 m/s
iii) a = v2/r
a = 2.71 x 10-3 m s-2
b) i) Change in GPE = Uo – Us, where o is orbit, s is surface
= m(ϕo – ϕs)
= 6.93 x 109 J
ii) By Newton’s second law of motion,
Fg = mrω2
Fg = mr(2π/T)2
T = 6566s
T = 1.82 hrs.
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iii)LEO satellites have a shorter orbital period as compared to geostationary satellites
which have a period of 24 hrs.
LEO satellites are nearer to the Earth and so the resolution of the images received will
be higher.
GRAVITATION NOTES H2 PHYSICS | Created by MS. Zheng, 91640715