Question 1
Main learning points
a. When we are considering the gravitational field strength outside the Earth, we should treat the mass of the Earth to be concentrated at its centre, hence becoming like a point mass.
b. Therefore, the equation g = GM/r^2 can only be used to determine the g-field strength outside the Earth (or above the Earth's surface).
c. For points within the Earth (or below the Earth's surface), the equation g = GM/r^2 can no longer be used. Instead, g becomes proportional to the distance away from the centre of the Earth.
d. The g-field strength inside a ring of mass is zero. Hence another mass which is placed inside the ring of mass, will not experience any gravitational force. This is similar to a ring of charged conductor (the charges in a conductor can be evenly distributed) where the electric field inside the ring will be zero - do not worry about this yet, you will learn more in the next topic on Electric Field. You can read up about Shell theorem but it is not within the syllabus.
Main learning points
a. When we are considering the gravitational field strength outside the Earth, we should treat the mass of the Earth to be concentrated at its centre, hence becoming like a point mass.
b. Therefore, the equation g = GM/r^2 can only be used to determine the g-field strength outside the Earth (or above the Earth's surface).
c. For points within the Earth (or below the Earth's surface), the equation g = GM/r^2 can no longer be used. Instead, g becomes proportional to the distance away from the centre of the Earth.
d. The g-field strength inside a ring of mass is zero. Hence another mass which is placed inside the ring of mass, will not experience any gravitational force. This is similar to a ring of charged conductor (the charges in a conductor can be evenly distributed) where the electric field inside the ring will be zero - do not worry about this yet, you will learn more in the next topic on Electric Field. You can read up about Shell theorem but it is not within the syllabus.
Question 2
Main learning points
a. A height of 144 km above Earth's surface would mean that it is of a distance of 144km + radius of Earth away from the centre of the Earth.
b. To prove that weight is reduced by 5%, it simply means (new weight) / (old weight) = 0.95. (Important!!!)
Main learning points
a. A height of 144 km above Earth's surface would mean that it is of a distance of 144km + radius of Earth away from the centre of the Earth.
b. To prove that weight is reduced by 5%, it simply means (new weight) / (old weight) = 0.95. (Important!!!)
Question 3
If you are unable to solve, please look at this first --> *Hint*
Main learning points
a. To determine the g-field strength at a point outside the sphere, you should consider the entire mass of the sphere to be at the centre of the mass.
b. The mass of the sphere with diameter 2d will be larger than the mass of the sphere of same density with diameter d.
Solution 1 (This is a continuation from the hint.)
Solution 2 (This is an alternative way in solving the question.)
If you are unable to solve, please look at this first --> *Hint*
Main learning points
a. To determine the g-field strength at a point outside the sphere, you should consider the entire mass of the sphere to be at the centre of the mass.
b. The mass of the sphere with diameter 2d will be larger than the mass of the sphere of same density with diameter d.
Solution 1 (This is a continuation from the hint.)
Solution 2 (This is an alternative way in solving the question.)
Question 4
Main learning points
a. The gravitational field strength is a vector quantity and has a direction, hence we will need to use vector diagram to determine the resultant g-field strength.
b. The vector sum of the gravitational field strengths at the neutral point is zero.
c. The neutral point always lies closer to the smaller mass.
Main learning points
a. The gravitational field strength is a vector quantity and has a direction, hence we will need to use vector diagram to determine the resultant g-field strength.
b. The vector sum of the gravitational field strengths at the neutral point is zero.
c. The neutral point always lies closer to the smaller mass.
Question 5
Main learning points
a. Be sure to read the question carefully and know exactly which values to use in the calculations.
b. It is proven in this question that the neutral point always lies closer to the smaller mass.
Main learning points
a. Be sure to read the question carefully and know exactly which values to use in the calculations.
b. It is proven in this question that the neutral point always lies closer to the smaller mass.
Question 6
Check your understanding of escape speed in this video --> Escape speed
Main learning points
a. Escape speed is the critical speed that the body must have during launching so that it is able to escape the Earth permanently.
b. Do not solve the problem simply by using the equation "Escape speed = (2GM/R)^1/2 as found in the lecture example. Marks may not be awarded if the solution is unclear. Remember, you have to show understanding rather than just mathematical solution.
c. Instead, have a good understanding that the KEmin must be equal or larger than the change in the GPE that the object is experiencing.
d. Escape speed is independent of the mass of the body.
--> Solution
Check your understanding of escape speed in this video --> Escape speed
Main learning points
a. Escape speed is the critical speed that the body must have during launching so that it is able to escape the Earth permanently.
b. Do not solve the problem simply by using the equation "Escape speed = (2GM/R)^1/2 as found in the lecture example. Marks may not be awarded if the solution is unclear. Remember, you have to show understanding rather than just mathematical solution.
c. Instead, have a good understanding that the KEmin must be equal or larger than the change in the GPE that the object is experiencing.
d. Escape speed is independent of the mass of the body.
--> Solution
Question 7
If you are unable to solve, do look at this video *hint*.
Main learning points
a. As long as the object is able to reach slightly on the right side of neutral point, it will be attracted towards the moon simply because the net force/g-field is towards the centre of the moon. It will be accelerated towards the centre of the moon the moment it crosses the neutral point.
b. The gravitational potential at the neutral point is not zero. Only the net g-field strength is zero. That means the GPE at the neutral point is not zero.
--> Solution
If you are unable to solve, do look at this video *hint*.
Main learning points
a. As long as the object is able to reach slightly on the right side of neutral point, it will be attracted towards the moon simply because the net force/g-field is towards the centre of the moon. It will be accelerated towards the centre of the moon the moment it crosses the neutral point.
b. The gravitational potential at the neutral point is not zero. Only the net g-field strength is zero. That means the GPE at the neutral point is not zero.
--> Solution
Question 8
Main learning point
a. Read the question carefully. Some of you may not have used the correct distances. One of the ways to prevent careless mistakes is to draw diagram while reading the question. Do not be over-confident.
Main learning point
a. Read the question carefully. Some of you may not have used the correct distances. One of the ways to prevent careless mistakes is to draw diagram while reading the question. Do not be over-confident.
Question 9
If you are unable to solve, do look at this video *hint*.
Main learning points
a. The total gravitation potential energy is the total work done required in bringing the components of the system (i.e. the moons and planets) from infinity and arrange these components individually into this configuration at some point in space.
b. For example, if we were to first bring the planet with mass M into some point in space, no work done is required as no external force is required. (The potential at that point in space is zero). However, after the mass M is in the space, (negative) work has to be done by external force to bring the first moon from infinity to the point. (The potential at a point around the mass M is now negative). To bring the second moon from infinity to the point, more (negative) work has to be done by external force. (The potential at the point is now the sum of the potential due to the mass M and the first moon). Likewise for the third moon.
--> Solution
If you are unable to solve, do look at this video *hint*.
Main learning points
a. The total gravitation potential energy is the total work done required in bringing the components of the system (i.e. the moons and planets) from infinity and arrange these components individually into this configuration at some point in space.
b. For example, if we were to first bring the planet with mass M into some point in space, no work done is required as no external force is required. (The potential at that point in space is zero). However, after the mass M is in the space, (negative) work has to be done by external force to bring the first moon from infinity to the point. (The potential at a point around the mass M is now negative). To bring the second moon from infinity to the point, more (negative) work has to be done by external force. (The potential at the point is now the sum of the potential due to the mass M and the first moon). Likewise for the third moon.
--> Solution
Question 10
Main learning points
a. The topic on gravitational field is closely linked to the topic of circular motion.
b. The net force acting on a body moving in uniform circular motion will always be pointing towards the centre of the circle.
c. Weight may not always equal to the centripetal force. You will need to look at the context and should always start from the free body diagram. For this question, the weight (i.e. gravitational force) equals to the centripetal force because the normal contact force acting on the object is zero.
Main learning points
a. The topic on gravitational field is closely linked to the topic of circular motion.
b. The net force acting on a body moving in uniform circular motion will always be pointing towards the centre of the circle.
c. Weight may not always equal to the centripetal force. You will need to look at the context and should always start from the free body diagram. For this question, the weight (i.e. gravitational force) equals to the centripetal force because the normal contact force acting on the object is zero.
Question 11
Solutions 11a, 11b, 11c, 11d
Main learning points
a. For an object which is orbiting around a mass, the net force acting on the object should equal to the gravitational force acting on it. The gravitational force provides the centripetal force required for the circular motion to take place.
b. When the mass of the Earth is not given in the question, it should be derived using the information given. You should not use a value which you have memorised, unless it is an estimation question.
c. Need to memorise the three conditions for a geostationary satellite as well as the reasons why these conditions must be fulfilled before the satellite is geostationary.
d. Need to show total energy of satellite = gravitational potential energy + kinetic energy instead of simply quote total energy = - kinetic energy.
e. The negative sign in the total energy of the satellite implies that the satellite does not have enough energy to escape the gravitational field of Earth, and is therefore bound to the Earth.
Solutions 11a, 11b, 11c, 11d
Main learning points
a. For an object which is orbiting around a mass, the net force acting on the object should equal to the gravitational force acting on it. The gravitational force provides the centripetal force required for the circular motion to take place.
b. When the mass of the Earth is not given in the question, it should be derived using the information given. You should not use a value which you have memorised, unless it is an estimation question.
c. Need to memorise the three conditions for a geostationary satellite as well as the reasons why these conditions must be fulfilled before the satellite is geostationary.
d. Need to show total energy of satellite = gravitational potential energy + kinetic energy instead of simply quote total energy = - kinetic energy.
e. The negative sign in the total energy of the satellite implies that the satellite does not have enough energy to escape the gravitational field of Earth, and is therefore bound to the Earth.
Question 12
(b)(i) Geostationary orbit is one orbit in which a satellite moves such that it is always positioned above a particular point on the Earth's surface.
(b)(ii) Simple calculations. You should have no problems solving. Please refer to this if you really need some help.
(c) Need to memorise the three conditions for a geostationary satellite as well as the reasons why these conditions must be fulfilled before the satellite is geostationary.
(d) Geostationary satellite allows signals to be transmitted steadily since its position is fixed with respect to the transmitting stations and receiving stations on Earth.
(b)(i) Geostationary orbit is one orbit in which a satellite moves such that it is always positioned above a particular point on the Earth's surface.
(b)(ii) Simple calculations. You should have no problems solving. Please refer to this if you really need some help.
(c) Need to memorise the three conditions for a geostationary satellite as well as the reasons why these conditions must be fulfilled before the satellite is geostationary.
(d) Geostationary satellite allows signals to be transmitted steadily since its position is fixed with respect to the transmitting stations and receiving stations on Earth.
Question 13
(a)(i) Angular velocity is defined as the rate of change of angular displacement of the object
(ii) Make use of angular velocity = 2 x pi / period. Period in this case is 365 days = 365 x 24 x 60 x 60 seconds.
(b)(i) You should have no problem solving for these two forces. You will need to make sure you have made use of the correct masses and distances.
(ii) You have to take note of the length of the arrows drawn. Make sure you use a ruler when drawing straight lines.
(iii) Resultant force is directed towards the centre of the sun. Acceleration = resultant force / mass of satellite.
(iv) The resultant force obtained in part (iii) is the centripetal force required for the circular motion of the satellite. The acceleration calculated in part (iii) is therefore the
centripetal acceleration of the satellite. The angular velocity should therefore be obtained by making use of the acceleration obtained in (iii) as it is the centripetal
acceleration.
(v) Watch the video for the explanation.
(vi) Disadvantages
1. The satellite is located closer to the Sun than Earth which means that it is constantly exposed to a higher temperature. The build of the satellite has to be able to
withstand this higher temperature and stronger light intensity, which could lead to higher manufacture costs and maintenance costs.
2. The satellite is much further away from the Earth (1.60 x 10^9 m) compared to a typical geostationary orbit (~10^7 m, refer to question 11). This could mean a higher
launching cost is required to launch this satellite.
3. The satellite is no longer positioned above a particular point on the Earth's surface as the Earth will rotate about it's own axis. This means that the data collected about
the Sun could not be sent to the data collection centre located at a position on Earth instantaneously.
(a)(i) Angular velocity is defined as the rate of change of angular displacement of the object
(ii) Make use of angular velocity = 2 x pi / period. Period in this case is 365 days = 365 x 24 x 60 x 60 seconds.
(b)(i) You should have no problem solving for these two forces. You will need to make sure you have made use of the correct masses and distances.
(ii) You have to take note of the length of the arrows drawn. Make sure you use a ruler when drawing straight lines.
(iii) Resultant force is directed towards the centre of the sun. Acceleration = resultant force / mass of satellite.
(iv) The resultant force obtained in part (iii) is the centripetal force required for the circular motion of the satellite. The acceleration calculated in part (iii) is therefore the
centripetal acceleration of the satellite. The angular velocity should therefore be obtained by making use of the acceleration obtained in (iii) as it is the centripetal
acceleration.
(v) Watch the video for the explanation.
(vi) Disadvantages
1. The satellite is located closer to the Sun than Earth which means that it is constantly exposed to a higher temperature. The build of the satellite has to be able to
withstand this higher temperature and stronger light intensity, which could lead to higher manufacture costs and maintenance costs.
2. The satellite is much further away from the Earth (1.60 x 10^9 m) compared to a typical geostationary orbit (~10^7 m, refer to question 11). This could mean a higher
launching cost is required to launch this satellite.
3. The satellite is no longer positioned above a particular point on the Earth's surface as the Earth will rotate about it's own axis. This means that the data collected about
the Sun could not be sent to the data collection centre located at a position on Earth instantaneously.
Question 14
(a)(i) You will need to explain your working. If you were to start your derivation with simply the equation GMm/r^2 = mv^2/r without explanation, you may not get the credit. Explanation in words is especially important when they ask you to explain your workings or when they ask you to "show that....".
(a)(ii) You should be able to show the relationship once you have derived the kinetic energy of the satellite correctly in part (i). Do not forget the negative sign.
(b)(i) When the question ask you to "draw" instead of "sketch", the points have to be plotted accurately on the graph. Marks will usually be credited for the accuracy of the points plotted.
(b)(ii) Please view the video for the common mistake that student make in the calculation. This applies to similar questions relating to the change in elastic potential energy in the spring.
Extension
Can b(ii) be solved by using the difference in kinetic energy = difference in gravitational potential energy? Or loss in gravitational potential energy = gain in kinetic energy?
The equation is based on conservation of energy. When can the equation be used then?
After watching the video, ask yourself, can the satellite move from one orbit to another without any input of energy? Should energy be taken away or input into the satellite in order to move the satellite from 4Rp to 2Rp?
(a)(i) You will need to explain your working. If you were to start your derivation with simply the equation GMm/r^2 = mv^2/r without explanation, you may not get the credit. Explanation in words is especially important when they ask you to explain your workings or when they ask you to "show that....".
(a)(ii) You should be able to show the relationship once you have derived the kinetic energy of the satellite correctly in part (i). Do not forget the negative sign.
(b)(i) When the question ask you to "draw" instead of "sketch", the points have to be plotted accurately on the graph. Marks will usually be credited for the accuracy of the points plotted.
(b)(ii) Please view the video for the common mistake that student make in the calculation. This applies to similar questions relating to the change in elastic potential energy in the spring.
Extension
Can b(ii) be solved by using the difference in kinetic energy = difference in gravitational potential energy? Or loss in gravitational potential energy = gain in kinetic energy?
The equation is based on conservation of energy. When can the equation be used then?
After watching the video, ask yourself, can the satellite move from one orbit to another without any input of energy? Should energy be taken away or input into the satellite in order to move the satellite from 4Rp to 2Rp?