Download Structured Questions (<8 marks)

Gravitation
Structured Questions (<8 marks)
(Level 1)
*Code: 13L1B001, Total marks: 3
Two bodies of the same of mass 2 × 1011 kg are 4 × 105 m apart.
(a) Find the gravitational force between them.
(2 marks)
(b) How does the answer in (a) change if the distance between the bodies is
doubled?
(1 mark)
Answer:
GMm
, the force is
r2
 2 1011  2 1011
(a) Applying F 
6.67 10  
 
4 10 
11
  16.675 N .
5 2
(b) The force will decrease to a quarter of the original value.
(1M+1A)
(1A)
*Code: 13L1B002, Total marks: 4
The top of Mount Everest (珠穆朗瑪峯) is h = 8850 m above the sea level.
The gravitational field strength at the sea level g0 is 9.81 N kg−1. The radius of the
Earth is r = 6.37 × 106 m.
(a) Find the gravitational field strength at the top of Mount Everest.
(2 marks)
(b) What is the difference in the weights between a man of 70 kg on the sea level
and another man of the same mass at the top of Mount Everest?
(2 marks)
Answer:
(a)
Applying g 
g  g0 
GM
, the gravitational field strength is
r2
r2
r  h 2
 9.81
6.37 10 
6 2
8850  6.37 10 
 9.783
 9.78 N kg 1
6 2
(1M+1A)
(b) The difference in the weights is 70 × (9.81 − 9.783) ≈ 1.90 N.
(1M+1A)
*Code: 13L1B003, Total marks: 4
A spacecraft is sent from the Earth to the Moon. At a certain position, the magnitude
of the gravitational force acting on the spacecraft due to the Earth is the same as that
due to the Moon.
(a) What is the ratio of the distance between the spacecraft and the Earth, rE to that
between the spacecraft and the Moon, rM?
(2 marks)
(b) Must the resultant force be zero at that position? If yes, explain your answer
briefly. If not, state the essential conditions for the resultant force to be zero.
(2 marks)
24
Mass of the Earth ME = 5.97 × 10 kg
Mass of the Moon MM = 7.35 × 1022 kg
Answer:
(a)
Let m be the mass of the spacecraft.
The gravitational force between two bodies is given by
GMm
. Hence
r2
GM E m GM M m

rE2
rM2
rE2 5.97  10 24

rM2 7.35  10 22
(1M+1A)
rE 9.01

rM
1
(b) The resultant force does not necessarily be zero (1A). It will be zero if the centre
of mass of the Earth, the spacecraft and the centre of mass of the Moon lie on a
straight line (1A).
*Code: 13L1B004, Total marks: 4
1
that of the Earth.
81
(a) What is the ratio of the gravitational force exerted on the Earth by the Moon to
that exerted on the Moon by the Earth?
(1 marks)
1
(b) The gravitational field strength on the surface of the Moon is
that of the
6
Earth. Compare the average density of the Moon and that of the Earth.
(3 marks)
The mass of the Moon is
Answer:
(a) The ratio is 1 : 1.
(1A)
(b) Let rE and rM be the radius of the Earth and that of the Moon. Let ME and MM be
the mass of the Earth and that of the Moon.
Consider the gravitational field strength.
GM M 1 GM E
  2
6
rM2
rE
 rE

 rM
2

81
 
6

(1M)
rE 3 6

rM
2
The average density of the Moon is
MM
ME
1 81


3
4 3  2 
4 3
πrM 
πrE



3
3
3 6 
ME
 0.6124 
4 3
πrE
3
The average density of the Moon is about 0.612 times that of the Earth.
(1M)
(1A)
*Code: 13L1B005, Total marks: 5
In the figure as shown, the Earth, the Moon and the Sun form a right-angled triangle,
with the Moon at the right angle.
Moon
Sun
Earth
(a)
What is the gravitational force acting on the Moon by the Sun?
(b)
What is the gravitational force acting on the Moon by the Earth?
(2 marks)
(1 mark)
(c) What is the resultant gravitational force (magnitude and direction) acting on the
Moon by the Sun and the Earth?
(2 marks)
Mass of the Sun = 1.99 × 1030 kg
Mass of the Earth = 5.97 × 1024 kg
Mass of the Moon = 7.35 × 1022 kg
Answer:



 

6.67 10 11  1.99 1030  7.35 10 22
GMm
(a) Applying F  2 , the force is
2
r
1.5 1011
= 4.336 × 1020 ≈ 4.34 × 1020 N.
(1M+1A)
11
24
22
6.67  10  5.97  10  7.35  10
(b) The force is
2
3.84  108
= 1.9848 × 1020 ≈ 1.98 × 1020 N.
(1A)

 


 


(c) The resultant force has a magnitude of
≈ 4.77 × 1020 N.
4.336 10   1.9848 10 
20 2
20 2
(1A)
 1.9848 10
The force points to the left in an angle of tan 1 
20
 4.336 10
line joining the Sun and the Moon.
20

 ≈ 24.6° below the

(1A)
*Code: 13L1B006, Total marks: 4
A planet revolves around a star in a circular orbit of a radius of 1.5 × 1011 m. The
mass of the planet and the star are 6 × 1024 kg and 2 × 1030 kg respectively.
(a) What is the gravitational force between the planet and the star?
(2 marks)
(b) What is the orbital period of the planet? Express your answer in terms of Earthyear.
(2 marks)
Answer:

 

 
6.67  10 11  2  10 30  6  10 24
GMm
(a) Applying F  2 , the force is
2
r
1.5  1011
22
22

= 3.557 × 10 ≈ 3.56 × 10 N.
(b) The period can be determined by
4π 2 3
T2 
r
GM
T
 
4π 2
 1.5  1011
6.67  10 11  2  1030

 

(1M+1A)

3
 3.160  10 7 s
3.160  10 7
The period is
≈ 1.00 Earth-year.
365  24  60  60
(1M)
(1A)
*Code: 13L1B007, Total marks: 4
Two satellites A and B are orbiting around a planet O in circular orbits. It is known
OA 2
 .
that
OB 3
(a) What is the ratio of the linear speed of A to that of B?
(2 marks)
(b) What is the ratio of the orbital period of A to that of B?
(2 marks)
Answer:
(a) Since v 
1
, the ratio of the linear speed of A to that of B is
r
3
≈ 1.22.
2
(b) Since T 2  r 3 , the ratio of the orbital period of A to that of B is
(1M+1A)
3
 22
  ≈ 0.544.
3
(1M+1A)
*Code: 13L1B008, Total marks: 4
An asteroid moves around the Sun in a circular orbit whose radius is 5 times that of
the Earth’s orbit.
(a) Find the orbital period of this asteroid.
(2 marks)
(b) Find the angular speed of this asteroid.
(2 marks)
Answer:
(a) The period of the Earth TE is one year.
Since T 2  r 3 , we have
T

 TE
2
 5
   
 1
T  11.18
The period is 11.2 Earth-years.
2π
(b) Applying  
, the angular speed is
T
2π
≈ 1.78 × 10−8 rad s−1
11.18  365  24  60  60
3
(1M)
(1A)
(1M+1A)
Structured Questions (<8 marks)
(Level 2)
*Code: 13L2B001, Total marks: 6
A spacecraft is set into a circular orbit around Jupiter. The radius of this orbit is 3.5
times the radius of the planet.
(a)
What is the period of the spacecraft in this orbit?
(3 marks)
(b) Another spacecraft is set into an orbit whose period is 1.2 times that of (a). What
is the orbital speed of this spacecraft?
(3 marks)
−1
The gravitational field strength on the surface of Jupiter is 23 N kg . The radius of
Jupiter is 71 500 km.
Answer:
(a) The gravitational field strength at a point in the orbit is
1
= 1.878 N kg−1.
(1M)
23
2
3.5
Consider the centripetal force acting on the spacecraft. The orbital speed is
mv 2
 mg
r
v2
 1.878
(1M)
3  7.15  10 7 
v  2.007  10 4 m s 1
2π  7.15  10 7
Therefore the period is
= 2.239 × 104 ≈ 2.24 × 104 s.
2.007  10 4
(b) Since T 2  r 3 , we have


(1M+1A)
3

r
1.2  
7 
 3  7.15 10 
r  2.422  108 m
2π  2.422 108
The speed is
≈ 5.67 × 104 m s−1.
1.2  2.239 10 4
2




(1M)


(1M+1A)
*Code: 13L2B002, Total marks: 6
A binary star system consists of two stars of the same mass 2 × 1030 kg which are
separated by a distance of 5 × 1010 m. The stars rotate about their common centre of
mass O and trace out a circle as shown.
(a) What is the gravitational force between the two stars?
(2 marks)
(b) Find the orbital speed and orbital period of the stars.
(4 marks)
Answer:

 

 
6.67  1011  2  1030  2  1030
GMm
,
the
force
is
2
r2
5  1010
= 1.067 × 1029 ≈ 1.07 × 1029 N.
mv 2
(b)
Applying F 
, the orbital speed is
r
2  1030  v 2
1.07  10 29 
2.5  1010
v  36 520
(a) Applying F 



(1M+1A)

(1M+1A)
1
 36 500 m s
2π  2.5  1010
The orbital period is
≈ 4.30 × 106 s.
36 520


(1M+1A)
Structured Questions (≥ 8 marks)
(Level 1)
*Code: 13L1C001, Total marks: 8
The Hubble Space Telescope has a mass of 11 110 kg. It orbits around the Earth at an
altitude of 559 km.
(a) What is the gravitational field at a point on the telescope’s orbit?
(2 marks)
(b) What is the centripetal force provided to the telescope?
(2 marks)
(c) What is the orbital speed of the telescope?
(2 marks)
(d) What is the orbital period of the telescope?
(2 marks)
24
Mass of the Earth = 5.97 × 10 kg
Radius of the Earth = 6370 km
Answer:
(a) The gravitational field strength is
GM 6.67 10 11  5.97 10 24

 8.294  8.29 N kg 1 .
2
2
6
5
r
6.37 10  5.59 10


 


(b) The centripetal force is 11 110 × 8.294 ≈ 9.21 × 104 N.
v2
(c) Applying a  , the speed is
r
v2
8.294 
6.37  10 6  5.59  10 5
v  7581
 7580 m s 1
(d) The orbital period is


2π  6.37 10 6  5.59 10 5
≈ 5740 s.
7581
(1M+1A)
(1M+1A)
(1M+1A)
Structured Questions (≥ 8 marks)
(Level 2)
*Code: 13L2C001, Total marks: 9
(a) A satellite is orbiting around the Earth in a circle. By considering the
gravitational force acting on the satellite by the Earth and the circular motion of
the satellite, show that its speed (in m s−1) is given by
2107
v
r
where r (in metres) is the distance between the centre of the Earth and the
satellite.
(3 marks)
(b) LAGEOS-1 satellite is a sphere of mass 411 kg. It is set into an orbit 5900 km
above the Earth’s surface.
(i)
What is the gravitational field strength at a point on the orbit?
(2 marks)
(ii) What is the gravitational force acting on the satellite by the Earth?
(2 marks)
(iii) What is orbital period of the satellite?
The mass of the Earth ME is 5.97 × 1024 kg.
The gravitational field strength on the Earth’s surface is 9.81 N kg−1.
The radius of the Earth is 6370 km.
(2 marks)
Answer:
(a) Let m be the mass of the satellite.
The gravitational force between the satellite and the Earth is
GM E m
.
r2
mv 2
.
r
Since the gravitational force provides the centripetal force, we have
GM E m mv 2

r
r2
GM E
v
r
The centripetal force provided to the satellite is
6.67 10  5.97 10 
11

24
(1A)
(1A)
(1A)
r
2  10
r
(b) The distance of the satellite from the centre of the Earth is
6370 + 5900 = 12 270 km or 1.227 × 107 m.
(i) The gravitational field strength is

7
2
 6370 
9.81 
 = 2.644 ≈ 2.64 N kg−1.
 12270 
(1M+1A)
(ii) The gravitational force is 411 × 2.644 = 1087 ≈ 1090 N.
(iii) From (a),
2  10 7
v
 5710 m s 1 .
7
1.227  10
2π  1.227  10 7
The orbital period is
≈ 13 500 s.
5710


(1M+1A)
(1M+1A)