Download Lecture 7 Gravity and satellites

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0430 860 810 Nick Zhang
Lecture 7 Gravity and satellites
Newton's Law of Universal Gravitation
Gravitation is a force of attraction that acts between any two masses.
The gravitation force acting between two masses, m1 and m2:
� is a force of attraction that acts from the centre of each mass
� acts equally on each mass (Newton's third law)
� is weaker if the masses are further apart. Gravitation is inversely proportional to the squared distance
between the centres of the masses
� depends directly on the mass of each body involved
Gm1m2
R2
F is the gravitational force acting on each body (N)
m1 and m2 are the masses of the objects (kg)
R is the distance between the centres of the objects (m)
Therefore, the universal gravitational force is given by: F =
where
The universal gravitational constant G is equal to 6.67 × 10−11 Nm 2 kg −2 .
Example
The gravitational force that acts on a 1000 kg space probe on the surface of Mars is 4.43 × 103 N. The
radius of Mars is 3400 km. Without using the mass of Mars, determine the gravitational force that acts on
the space probe when it is:
a. 3400 km above the surface of Mars.
b. 6800 km above the surface of Mars.
Gravitational field strength
A gravitational field is a region in which any object will experience a gravitational force.
The magnitude of a gravitational field is known as the gravitational field strength, g. This is defined as the
GM
gravitational force that acts on each kilogram of an object in the field and is given by: g = 2
R
where
G is the universal gravitational constant 6.67 × 10−11 Nm 2 kg −2
M is the mass of the central planet (kg)
R is the distance from the centre of the central planet (m)
g is the gravitational field strength (N/kg)
NOTE that the gravitation field strength does not depend on the mass of the object in the gravitational field.
The gravitational force acting on an object in a gravitational field is also called the weight, W, of the object
GMm
and is given by: W = mg =
R2
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0430 860 810 Nick Zhang
Example
a. Calculate the gravitational field strength g at the surface of the Earth (mass of Earth is 6.0 × 1024 kg ,
radius of Earth is 6.4 × 106 m ).
b. Calculate the gravitational field strength g at the surface of the Moon (mass of Moon is 7.34 × 1022 kg ,
radius of Moon is 1.74 ×106 m ).
c. Determine the weight of an astronaut, whose total mass is 100 kg, at each of these locations.
Example
The planet Alpha, whose mass is M, has one moon Beta of mass 0.01M. The mean distance between the
centres of Alpha and Beta is R.
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a. If an asteroid is at point X, exactly halfway between the centres of Alpha and Beta, calculate the value of
the ratio:
force exerted on asteroid by Alpha
force exerted on asteroid by Beta
b. Point Y represents the distance from planet Alpha where the magnitude of the net gravitational force is
zero. What is this distance in terms of R?
A broader concept of free falling
As long as an object is accelerating towards the centre of the central planet with an acceleration equal to the
gravitational field strength at that location, the motion of the object could be considered as free falling. A
person during free falling would experience apparent weightlessness (a = g and N = 0)
Motion of the planets
A satellite is an object that is in a stable orbit around a more massive central planet. The only force acting on
a satellite is the gravitational attraction between it and the central planet. Satellites are in continuous free-fall
(uniform circular motion) with the centripetal acceleration being equal to the gravitational field strength at
the location of their orbit.
For the satellite orbiting around the central
planet, the gravitational force exerted by the
central planet on the satellite provides the
centripetal force.
F
G is the universal gravitational constant
GMm
F=
= ma
R2
⎧ v2
⎪⎪
a = ⎨ R2
⎪ 4π R
⎪⎩ T 2
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6.67 × 10−11 Nm 2 kg −2
GM
=a
R2
a=g
M is the mass of the central planet (kg)
R is the distance from the centre of the central
g=
GM
R2
planet (m)
g is the gravitational field strength (N/kg)
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a=g
a=g
v 2 GM
= 2
R
R
v2 =
GM
R
v=
GM
R
0430 860 810 Nick Zhang
4π 2 R GM
= 2
T2
R
4π 2 R3 = GMT 2
(orbital speed)
Kepler's third law
R 3 GM
=
T 2 4π 2
Example
Assume that somewhere in space there is a small spherical planet with a radius of 30 km. By some chance a
person living on this planet visits Earth. He finds that he weighs the same on Earth as he did on his home
planet, even though Earth is so much larger.
Earth has a radius of 6.37 × 106 m and a mass of 5.98 × 1024 kg.
The acceleration due to gravity (g), or the gravitational field, at the surface of Earth, is approximately 10
N/kg. The universal gravitational constant, G, is 6.67 × 10−11 Nm 2 kg −2 .
What is the value of the gravitational field on the surface of the visitor's planet?
What is the mass of the visitor's planet? Explain your answer by showing clear working.
The visitor's home planet is in orbit around its own small star at a radius of orbit of 1.0 × 109 m. The star
has a mass of 5.7 × 1025 kg.
What would be the period of the orbit of the visitor's planet? Show working.
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Example
Nick is a geostationary satellite. Its period of orbit is 24 hours so that it revolves at the same rate at which
the Earth turns. Given that the mass of the Earth is 6.0 × 1024 kg and the mass of Nick is 1200 kg, calculate:
a. its orbital radius
b. the gravitational field strength at this radius
c. Nick's orbital speed
d. its acceleration.
Example
Ceres, the first asteroid to be discovered, was found in 1801. Ceres has a mass of 7.0 × 1020 kg and a radius
of 385 km.
a. What is the gravitational field strength at the surface of Ceres?
b. Determine the speed required by a satellite in order to remain in orbit 10 km above the surface of Ceres.
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Example
The dwarf planet Pluto was discovered in 1930, and was thought to be the outermost member of our solar
system. It can be considered to orbit the Sun in a circle of radius 6 billion kilometres ( 6.0 × 1012 m). In 2003
a new dwarf planet, Eris, was discovered. It has approximately the same mass as Pluto, but the average
radius of its orbit around the Sun is 10.5 billion kilometres ( 10.5 ×1012 m).
Which of the choices (A-D) below gives the best estimate of the ratio
gravitational attraction of the Sun on Eris
?
gravitational attraction of the Sun on Pluto
A. 0.33
B. 0.57
C. 1.75
D. 3.06
The period of Pluto around the Sun is 248 Earth-years.
How many Earth-years does Eris take to orbit the Sun?
Example
The planet Mars has a mass of 6.4 × 1023 kg, which is approximately
1
that of Earth, and its radius is
10
approximately half that of Earth.
Which of the following (A-D) gives the best value for the acceleration due to gravity at the surface of Mars?
A. 1 m/s2
B. 2.5 m/s2
C. 4 m/s2
D. 5 m/s2
The Mars probe that was launched in August 2005 is now orbiting Mars in an orbit with an average radius of
3.00 × 107 m.
What is the period of the orbit in seconds?
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