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Lesson 18 questions – Gravitational Fields
1
A binary star is a pair of stars that move in circular orbits around their
common centre of mass. For stars of equal mass, they move is the same circular orbit,
shown by the dotted line in the diagram. In this question, consider the stars to be point
masses situated at their centres at opposite ands of a diameter of the orbit.
a)
each star.
i)
Draw on the diagram arrows to represent the force acting on
(2)
ii)
Explain why the stars must be diametrically opposite to travel
in the circular orbit.
……For circular motion there must be centripetal force
……This force is toward the centre of the circle – in this case the attraction due to
gravity and so must be along the diameter of the circle making the planets opposite
each other.…………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
Newton’s law of gravitation applied to the situation in the diagram
may be expressed as
F = GM2/4R2
State what each of the symbols listed below represent
F ……force of attraction between Masses/stars……………………………
G ……gravitational constant…………………………………………
M ……mass of a star……………………………………………………
R ……radius of orbit…………………………………………………… (2)(1 for 2 2
for 4)
c)
i)
Show that the orbital period T of each star is related to its speed
v by v=2πR/T.
v=distance/time = 2πR/T
(1)
ii)
Show that the magnitude of the centripetal force required to
keep each star moving in its circular path is
F = 4π 2MR/T2
F = mv 2/r
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v=2πR/T
F = m(2πR/T)2/r
F = 4π 2MR/T2
iii)
(2)
Use equations from (b) and (ii) above to show that the mass of
each star is given by
M = 16π 2R3/GT2
F = GM2/4R2
F = 4π 2MR/T2
2
4π MR/T2= GM2/4R2
M = 16π 2R3/GT2
(2)
d)
Binary stars separated by a distance of 1x1011m have been observed
with an orbital period of 100 days. Calculate the mass of each star.
1 day = 86400s
M = 16π 2R3/GT2
M = (16π 2(0.5x1011)3)/((6.6710-11)86400002)
Mass = ……4.0 x 1030……………..kg (2)
2
This question is about gravitational fields. You may assume that all the mass
of the Earth, or the Moon, can be considered as a point mass at its centre.
a)
It is possible to find the mass of a planet by measuring the gravitational
field strength at the surface of the planet and knowing its radius.
i)
Define gravitational field strength, g.
………Gravitational force per unit mass…………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
ii)
Write down an expression for g at the surface of a planet in
terms of its mass M and its radius R.
g=GM/R2
(1)
iv)
M=g R2/G
24
Show that the mass of the Earth is 6.0x10 kg.
Radius of Earth = 6400km.
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M=9.81x 64002/6.6710-11
M=6.0x1024kg
(1)
b)
i)
Use the data below to show the value of g at the Moon’s
surface is about 1.7 Nkg-1.
Mass of Earth = 81 x mass of Moon
Radius of Earth = 3.7 x radius of Moon
2
g=GM/R
g=6.6710-11x(6.0x1024/81)/(6400000/3.7) 2
g=1.65 Nkg-1.
(2)
ii)
Explain why a high jumper who can clear a 2m bar on Earth
should be able to clear a 7m bar on the Moon. Assume that the high jump on the
Moon takes place inside a “space bubble” where Earth’s atmospheric conditions exist.
………ΔgE/ ΔgM = ΔhE/ ΔhM =5.8……………………………………………………..
………If high-jumoers centre of mass is at 1m
So high jumper lifts centre of mass 1m on earth to reach 2m…………………..
………On the Moon raises it 5.8m to make 6.8m
…………………………………………………………………………………………..
……………………………………………………………………………………… (3)
c)
The distance between the centres of the Earth and the Moon is 3.8 x
108 m. Assume that the moon moves in a circular orbit about the centre of the Earth.
Estimate the period of this orbit to the nearest day.
Mass of Earth = 6.0x1024kg
1 day = 86400s
F=mv2/R; F = m(2πR/T)2/R= GMm/ R2
(2πR/T)2/R= GM/ R2
T2=4π2R3/GM
22
M is mass of moon = 7.4 x 10
R = 3.8 x 108 m
Period = …………27…………… day (5)
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