Category Archives: A-level Units

Gravitation Quiz

These are 10 multiple choice questions on the topic of Gravitation according to the Singapore-Cambridge A-level Physics exams (syllabus 9646). Try it out to test your understanding of the topic. Worked solutions are provided at the end of the quiz.

Gravitation

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Question 1
A boy can jump to a maximum height of h on the surface of Earth. If he were on another planet whose radius is four times that of Earth’s, and density is only half that of Earth’s, determine the maximum height the boy can attain when he jumps.
A
\frac{h}{4}
B
\frac{h}{2}
C
2h
D
4h
Question 1 Explanation: 
\frac{h}{2} is the answer as g=\frac{GM}{r^2} \\= \frac{G\rho{\frac{4}{3}}\pi r^3}{r^2} \\=  G\rho{\frac{4}{3}}\pi r

As \rho decreases by half and r increases to 4 times, the overall effect is that g is doubled.

Hence, if we approximate the change in gravitational potential energy as the boy jumps to be mgh, his maximum height is halved.

Question 2
An Earth-orbiting satellite experiences a continuous small resistance from external influences and gradually transits into lower orbits. Which of the following correctly gives the changes in the kinetic energy and the gravitational potential energy of the satellite?
A
Kinetic energy increases and gravitational potential energy increases
B
Kinetic energy increases and gravitational potential energy decreases
C
Kinetic energy decreases and gravitational potential energy increases
D
Kinetic energy decreases and gravitational potential energy decreases
Question 2 Explanation: 
From the graph below, for an object in orbit (where gravitational force provides the centripetal force), when distance from the centre of the orbit r decreases, kinetic energy E_k increases and potential energy U decreases.

Question 3
Star X of mass 2M and Star Y of mass M perform circular motion about their common centre of mass under their gravitational attraction. What is the ratio of the force acting on X to the force acting on Y, ignoring the effects of any other bodies?
A
0.5
B
1
C
2
D
4
Question 3 Explanation: 
There is no need to do any calculations as the forces are action-reaction pairs. This is a tricky one.
Question 4
The diagram below shows how the gravitational potential varies between the moon and Earth. At which position will a particle experience zero net force?
A
A
B
B
C
C
D
D
Question 4 Explanation: 
Since F = mg and g=-\frac{\text{d}\phi}{\text{d}r}, the position with zero net force is when the gradient of the \phi-r graph is zero.
Question 5
Two stars of equal mass M move with constant speed v in a circular orbit of radius R about their common centre of mass as shown in the diagram below.

Binary Stars

What is the net force on each star?

A
\frac{GM^2}{4R^2}
B
\frac{Mv^2}{2R}
C
zero
D
\frac{2Mv^2}{R}
Question 5 Explanation: 
The distance apart is 2R. Using Newton's law of gravitation, F=\frac{GM^2}{(2R)^2}. Alternatively, F=\frac{Mv^2}{R} could have been an option too as the gravitational force provides the centripetal force.
Question 6
The acceleration of free-fall at the Earth’s surface is g and the radius of the Earth is R. What will be the acceleration of free fall at a height h above the surface?
A
\frac{gR^2}{h^2}
B
\frac{gh^2}{R^2}
C
\frac{gh}{R+h}
D
\frac{gR^2}{(R+h)^2}
Question 6 Explanation: 

g\propto{\frac{1}{R^2}}

g'(R+h)^2=gR^2

Question 7
Two satellites A and B orbit the Earth in circular orbits, the radius of satellite A’s orbit being 4 times that of satellite B. If the orbital period and tangential velocity of satellite A are T and v respectively, what are the corresponding values for satellite B?
Period Tangential velocity
A 8T 2v
B \frac{1}{8}T 2v
C \frac{1}{8}T \frac{1}{2}v
D 8T \frac{1}{2}v
A
A
B
B
C
C
D
D
Question 7 Explanation: 

F=\frac{GMm}{r^2}=mr{\omega}^2 \\
\frac{GM}{r^3}=({\frac{2\pi}{T}})^2 \\
r^3 \propto T^2

When r_B = \frac{1}{4}r_A, T_B = \frac{1}{8}T_A

F=\frac{GMm}{r^2}=\frac{mv^2}{r} \\
r \propto \frac{1}{v^2}

When r_B = \frac{1}{4}r_A, v_B = 2v_A

Question 8
The diagram shows two points A and B which are at distances X and 2X from the centre of the Earth respectively. At point A, the gravitational potential is -8 kJ kg-1.

What is the change in potential energy when a satellite of mass 2.0 kg is moved from A to B?
A
- 4 kJ
B
- 8 kJ
C
+ 4 kJ
D
+ 8 kJ
Question 8 Explanation: 
\phi_B = - 4 \text{ kJ kg}^{-1} \\
\Delta \phi= \phi_B - \phi_A \\
= +4 \text{ kJ kg}^{-1} \\
\Delta U = m \Delta \phi \\
= +8 \text{ kJ}
Question 9
A satellite is at a height h above the surface of the Earth. If the radius of the Earth is r, and the acceleration due to gravity at the Earth’s surface is g, the period of orbit of the satellite will be
A
2\pi\sqrt{\frac{r+h}{g}}
B
2\pi\sqrt{\frac{(r+h)^2}{gr}}
C
2\pi\sqrt{\frac{r^2}{g(r+h)}}
D
2\pi\sqrt{\frac{(r+h)^3}{gr^2}}
Question 9 Explanation: 
g=\frac{GM}{r^2} \\
g'=(\frac{r}{r+h})^2 g \\
g' = (r+h)\omega^2=(r+h)(\frac{2\pi}{T})^2 \\
T =2\pi\sqrt{\frac{(r+h)^3}{gr^2}}
Question 10
An artificial satellite is orbiting around the Earth at a steady speed. The astronaut in the satellite can be regarded as ‘weightless’ because
A
the gravitational force acting on him is zero.
B
the centripetal force he experienced is zero.
C
the satellite’s acceleration is the same as his.
D
the acceleration he experienced is zero.
Question 10 Explanation: 
When the satellite's acceleration is the same as his, all of the gravitational force acting on him is exactly providing for the centripetal force, hence there is no contact between him and the floor of the satellite.
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Definition of the Tesla

I was just discussing the different definitions that some JCs give for the definition of the Tesla (the SI unit for magnetic flux density) with my JC2 students last week. According to the 'A' level examiners, they are looking for something more like:

The magnetic flux density of a magnetic field is said to be 1 Tesla, if the force acting per unit length on an infinitely long conductor carrying a current of 1 A and placed perpendicularly to the magnetic field is 1 N m-1.

Rather than what is found in some A-level guide books:

A tesla is the magnetic flux density if a force of 1 N acts on a wire of length 1m, carrying a current of 1A placed perpendicular to the magnetic field.

Some JCs still teach this erroneous definition in their lectures.

So, why is the second definition wrong? In the measurement of a tesla, if we refer only to a wire of finite length, due to the end effects, the magnetic flux density at both ends of the wire will not be the same as that in the middle.

I will be addressing this error and more in my upcoming June holiday Physics intensive workshop meant for JC1 and JC2 students. Do sign up early to avoid disappointment!

Cartesian diver

The Cartesian diver is a simple toy that you can build at home using the simplest items such as a pen cap and plastic bottle. Using this toy, you can learn about upthrust and buoyancy.

When the plastic bottle is squeezed, the pressure compresses the air in the pen cap and the amount of upthrust decreases until it is less than the weight of the pen cap as less water is displaced. Hence, the pen cap sinks.

When it is released, upthrust increases to its original amount again and the pen cap floats again.

Follow this link http://www.scienses.com/cartesian-diver/ for a detailed explanation and video demonstration. It's done by my good friend who is also a JC physics lecturer!