__Introduction:__

Hi, I'm Michael, and I'm aÂ second year physicist at Durham. I love my subject and I've really enjoyed studying it further at University. I have previously tutored students in Maths and I've also helped teach English to asylum seekers who had no prior English skills. From these experiences, I have developed the patience to work with beginners and people who have a basic understanding but may be struggling with certain areas.

__The Tutorials__

The content of the tutorials will be dictated by you so that we can focus on areas which you may be struggling with. I believe that an understanding of the subject matter is essential and underpins exam technique as well as providing confidence when answering questions. From this understanding you can start to dissect exam questions and break them down into easy to manage steps. Examples are a great way of demonstrating the process of answering questions and I will use them to help tackle areas which you may be struggling with. I will aim to make the tutorials enjoyable and engaging so that we can make the most of the 55 minute slot.

__About Me__

I am very passionate about Physics. At school, I started the Physics society and arranged a school trip to a nearby particle accelerator. Maths goes hand in hand with Physics and many of the skills that I learnt at GCSE and A-level I now use on a daily basis as part of my degree. At school, I achieved all A star grades in my GCSEs and at A-level where I studied Maths, Physics and Chemistry. I also taught myself AS-level Further Maths and I hope to convey some of the techniques required to learn independently outside the classroom.

__What Next?__

If you have any questions, don't hesitate to ask them throughÂ a messageÂ or by booking a 'Meet-the-Tutor-Session' and I will get back to you as soon as possible. Please let me know which exam board you are studying with and also which areas you are having trouble with so that I can prepare appropriately.

I look forward to meeting you!

__Introduction:__

Hi, I'm Michael, and I'm aÂ second year physicist at Durham. I love my subject and I've really enjoyed studying it further at University. I have previously tutored students in Maths and I've also helped teach English to asylum seekers who had no prior English skills. From these experiences, I have developed the patience to work with beginners and people who have a basic understanding but may be struggling with certain areas.

__The Tutorials__

The content of the tutorials will be dictated by you so that we can focus on areas which you may be struggling with. I believe that an understanding of the subject matter is essential and underpins exam technique as well as providing confidence when answering questions. From this understanding you can start to dissect exam questions and break them down into easy to manage steps. Examples are a great way of demonstrating the process of answering questions and I will use them to help tackle areas which you may be struggling with. I will aim to make the tutorials enjoyable and engaging so that we can make the most of the 55 minute slot.

__About Me__

I am very passionate about Physics. At school, I started the Physics society and arranged a school trip to a nearby particle accelerator. Maths goes hand in hand with Physics and many of the skills that I learnt at GCSE and A-level I now use on a daily basis as part of my degree. At school, I achieved all A star grades in my GCSEs and at A-level where I studied Maths, Physics and Chemistry. I also taught myself AS-level Further Maths and I hope to convey some of the techniques required to learn independently outside the classroom.

__What Next?__

If you have any questions, don't hesitate to ask them throughÂ a messageÂ or by booking a 'Meet-the-Tutor-Session' and I will get back to you as soon as possible. Please let me know which exam board you are studying with and also which areas you are having trouble with so that I can prepare appropriately.

I look forward to meeting you!

No DBS Check

5from 6 customer reviews

Mohamed (Student)

May 23 2015

Excellent teacher

Mohamed (Student)

May 30 2015

Very happy with the help received.

Mohamed (Student)

June 7 2015

Very good tutor, highly recommended,

Mohamed (Student)

June 17 2015

Very good explanations, good all round teacher. Highly recommended.

As you will be aware, it is not easy to directly integrate terms involving sin^2(x) and cos^2(x), so we use a substitution to turn them into something which we can integrate.

From the double angle formula for cosine we have:

cos^2(x)=(cos(2x)+1)/2

sin^2(x)=(1-cos(2x))/2

This substitution removes the terms of sin^2(x) and cos^2(x) which we can't integrate and replaces them with terms involving cos(2x) which integrates to sin(2x)/2, which we know from integration by inspection.

Now consider the example above, integrate (1+sin(x))^2:

To start this question we need to expand the brackets so that we can integrate each term individually. (1+sin(x))^2 expands to 1+2sin(x)+sin^2(x).

Now we have three separate terms which we can integrate:

1 integrates to x

2sin(x) integrates to -2cos(x)

Now substitute sin^2(x) for 1/2-cos(2x)/2. This integrates to x/2-sin(2x)/4

To get our final answer we now add all of the terms we've just integrated together. Remember to include the constant of integration c.

x-2cos(x)+x/2-sin(2x)/4+c

this simplifies to:

3x/2-2cos(x)-sin(2x)/4+c

This is the final answer for this question.

As you will be aware, it is not easy to directly integrate terms involving sin^2(x) and cos^2(x), so we use a substitution to turn them into something which we can integrate.

From the double angle formula for cosine we have:

cos^2(x)=(cos(2x)+1)/2

sin^2(x)=(1-cos(2x))/2

This substitution removes the terms of sin^2(x) and cos^2(x) which we can't integrate and replaces them with terms involving cos(2x) which integrates to sin(2x)/2, which we know from integration by inspection.

Now consider the example above, integrate (1+sin(x))^2:

To start this question we need to expand the brackets so that we can integrate each term individually. (1+sin(x))^2 expands to 1+2sin(x)+sin^2(x).

Now we have three separate terms which we can integrate:

1 integrates to x

2sin(x) integrates to -2cos(x)

Now substitute sin^2(x) for 1/2-cos(2x)/2. This integrates to x/2-sin(2x)/4

To get our final answer we now add all of the terms we've just integrated together. Remember to include the constant of integration c.

x-2cos(x)+x/2-sin(2x)/4+c

this simplifies to:

3x/2-2cos(x)-sin(2x)/4+c

This is the final answer for this question.

The first part of this question relies on remembering one key principle which is that for a satellite, or any body orbiting in a circle, centripetal force is provided by gravitational force. The rest is simply algebraic manipulation as long as you can remember the equations for circular motion and gravitational force.

So, begin by equating centripetal force and gravitational force:

mv^2/r=GMm/r^2

You can see immediately that m, which is the mass of the satelite, is on both sides of the equaiton and cancels out. As this term is not in the final equation we are aiming for, this is a good start. You can also see that r is on both sides of the equation so we want to gather these together by multiplying both sides by r, leaving:

v^2=GM/r

This cannot be simplified further, however v, which is the speed of the satellite, is not in the final equation. We know that the speed of an object in circular motion is given by the equation v=2pir/T, where T is the period of the satellite's orbit. Substituting this into v^2=GM/r gives:

4pi^2r^2/T^2=GM/r

If we rearrange for T^2 we get:

T^2=(4pi^2r^3)/GM

which is what the question asked us to find. This is an important equation which can be use to find many features of an orbit and it is used regularly so it is important to learn the above derivation. It's not a difficult derivation with only two key steps, centripetal force=gravitational force and substituting v for v=2pir/T.

The second part of the question requires you to use the equation which we just derived. As when using any equation it is worthwhile checking which values you know and which you need to calculate:

T=Orbital period of a geostationary satellite=24 hours=8.64*10^4 seconds

M=Mass of the Earth 6*10^24kg

G=Gravitational constant=6.67*10^-11 Nm^2/kg^2

pi=3.14 or use calculator value

r= radius of the satellite's orbit, what we are trying to find

We now know all the terms in the equation apart from the one which we wish to calculate. So we can rearrange the equation for r and substitute in the above values. Also note that values of M, G and the number of seconds in a day will be provided in the data sheet in the exam and so don't need to be memorised. We then have:

r^3=(T^2GM)/(4pi^2)

Substituting in the values above we find:

r^3=((8.64*10^4)^2*6.67*10^-11*6*10^24)/(4*3.14^2)=7.57*10^22

Then take the cubic root to find r

r=4.23*10^7 m

This is the orbital radius of any geostationary satellite as we can see that it does not depend on the mass of the satellite, only the mass of the Earth. Also remember that geostationary satellites orbit above the equator and have an orbital period of 1 day.

The first part of this question relies on remembering one key principle which is that for a satellite, or any body orbiting in a circle, centripetal force is provided by gravitational force. The rest is simply algebraic manipulation as long as you can remember the equations for circular motion and gravitational force.

So, begin by equating centripetal force and gravitational force:

mv^2/r=GMm/r^2

You can see immediately that m, which is the mass of the satelite, is on both sides of the equaiton and cancels out. As this term is not in the final equation we are aiming for, this is a good start. You can also see that r is on both sides of the equation so we want to gather these together by multiplying both sides by r, leaving:

v^2=GM/r

This cannot be simplified further, however v, which is the speed of the satellite, is not in the final equation. We know that the speed of an object in circular motion is given by the equation v=2pir/T, where T is the period of the satellite's orbit. Substituting this into v^2=GM/r gives:

4pi^2r^2/T^2=GM/r

If we rearrange for T^2 we get:

T^2=(4pi^2r^3)/GM

which is what the question asked us to find. This is an important equation which can be use to find many features of an orbit and it is used regularly so it is important to learn the above derivation. It's not a difficult derivation with only two key steps, centripetal force=gravitational force and substituting v for v=2pir/T.

The second part of the question requires you to use the equation which we just derived. As when using any equation it is worthwhile checking which values you know and which you need to calculate:

T=Orbital period of a geostationary satellite=24 hours=8.64*10^4 seconds

M=Mass of the Earth 6*10^24kg

G=Gravitational constant=6.67*10^-11 Nm^2/kg^2

pi=3.14 or use calculator value

r= radius of the satellite's orbit, what we are trying to find

We now know all the terms in the equation apart from the one which we wish to calculate. So we can rearrange the equation for r and substitute in the above values. Also note that values of M, G and the number of seconds in a day will be provided in the data sheet in the exam and so don't need to be memorised. We then have:

r^3=(T^2GM)/(4pi^2)

Substituting in the values above we find:

r^3=((8.64*10^4)^2*6.67*10^-11*6*10^24)/(4*3.14^2)=7.57*10^22

Then take the cubic root to find r

r=4.23*10^7 m

This is the orbital radius of any geostationary satellite as we can see that it does not depend on the mass of the satellite, only the mass of the Earth. Also remember that geostationary satellites orbit above the equator and have an orbital period of 1 day.