INNOVA JUNIOR COLLEGE JC 1 H2 Physics (2012) - ASKnLearn

INNOVA JUNIOR COLLEGE
JC 1 H2 Physics (2012)
15
Name: ( ) Class: ________
FA8: Gravitational Field
Duration: 25 minutes
Data
Gravitational constant, G = 6.67 × 10 -11 N m 2 kg -2
Formula
Gravitational Potential, = GM
r
1 The figure below shows a planet P of mass m, orbiting the Sun S of mass M in a circular path of radius R.
S
R
P
(a) Write down an expression, in terms of G, m, M and R for the force exerted by the Sun on the planet.
[1]
GMm
FG
[1]
2
R
GM
(b) Show that the expression for the angular velocity of the planet in its orbit is ω =
3
R
The gravitational force provides the centripetal force. [1]
F G = F C
GMm
2
2 mR
[1]
R
GM
3
R
[2]
(c) The Earth is 1.5 ×10 11 m from the centre of the Sun and takes exactly one year to complete one orbit.
The planet Jupiter takes 11.9 years to complete an orbit of the Sun. Using the expression in (b),
calculate the radius of Jupiter’s orbit.
2
From (b),
3
T R
2
2 GM
3
T
E
R
E
3
T R
TJ
RJ
2
3
[1]
11
1 1.510
11.9 RJ
[1]
11
R 7.8 10 m
[1]
J
radius = ………………………………… m [3]
1

2 A satellite of mass 20 kg orbits the Earth with a uniform circular motion of period 24 hours, so that it is
always directly above the same point on the Earth.
(a) Explain why the orbit of the satellite must lie in the equatorial plane.
The Earth is rotating from West to East about its axis. If it’s not placed at the equatorial plane, it will
be sometimes over the northern or southern hemisphere [1] and hence not directly above the same
point on the Earth.
Or
The gravitational force on the satellite acts towards the centre of the Earth. The satellite needs to be
placed in a horizontal plane that cuts through the centre of the Earth and yet rotate in the same
direction as the Earth’s [1], hence it has to be the equatorial plane.
(b) Given that the gravitational field strength on the Earth’s surface is 9.81 m s -1 and the radius of the
Earth is 6.4 ×10 6 m, calculate
(i) the radius of the orbit
F G = F C
GMm 2
2 mr
r T
2
GM 2
3
r T
gr E
3
r
2
2
T
2
6 2
(9.81)(6.4 10 ) 2
r
3
2
246060
2
[1]
[1]
r = 4.2 × 10 7 m [1]
radius = ………………………………… m [3]
(ii)
the satellite’s gravitational potential energy,
U
GMm
r
U
gR m
r
(9.81)(6.410 ) (20)
4.2 10
[1]
2 6 2
E
7
U
8
1.91310 J
U
8
1.910 J
[1]
gravitational potential energy = ………………………………… J [2]
2

(ii) the satellite’s kinetic energy,
GMm
K
2r
2 6 2
gRE
m (9.81)(6.410 ) (20)
K
7
2r
2(4.2 10 )
[1]
7
K 9.567
10 J
7
K 9.6 10 J
[1]
kinetic energy = ………………………………… J [2]
(iii) the satellite’s total energy.
8 7
E ( 1.91310 ) (9.56710 )
T
E
T
7
9.610 J
[1]
total energy = ………………………………… J [1]
END OF PAPER
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