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Degree: PhD in Astronomy (Doctorate)  Cambridge University
About Me:
I'm a recent Physics graduate of Oxford University (1st Class Honours), currently working on a PhD in Astronomy at Cambridge University. My Astrophysics research is focused on characterising the atmospheres of planets orbiting other stars, in the hopes of one day detecting signs of life outside the solar system.
I have a lifelong passion for all things related to Science, which I'm always keen to share as a guest lecturer in schools around the UK and abroad!
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I have extensive experience in Science and Maths tutoring (particularly in Physics, which I tutor to undergraduates at Cambridge). My priority is not just to help you understand science on a deep level, but to appreciate why the natural world is so incredible!
I customise my tutoring to your individual needs, based on an approach of teaching by inquiry driven learning (as is the standard practise in Cambridge and Oxford). If you're looking for a deeper understanding than simply memorising equations to pass exams, that's exactly what I strive to deliver!
Oxbridge /University Applicants:
Having gone to both Oxford and Cambridge, I am more than happy to discuss applying to these universities, UCAS more generally and give advice on preparing for entrance exams and university as a whole.
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The key point to recognise here is that an object undergoing a circular orbit around the Earth is undergoing circular motion. Circular motion is caused by a centripetal force (acting towards the centre of the circle), described by:
F_{c }= m_{satellite}v^{2}/r (eq 1)
In the case of an orbit, the force acting towards the centre is due to the gravity of the Earth, which is described by Newton's law of universal gravitation:
F_{g }= Gm_{Earth}m_{Satellite}/r^{2 }(eq 2)
We can now equate these two expressions:
F_{c} = F_{g} => m_{satellite}v^{2}/r = Gm_{Earth}m_{Satellite}/r^{2 }(eq 3)
Now me can rearrange to make v the subject of the equation:
v = sqrt(Gm_{Earth}/r) (eq 4)
This is the speed of a satellite in a circular orbit around the Earth. Note that v is proportional to 1/sqrt(r) and that the mass of the satellite has cancelled out.
see moreThis is a quick derivation of the quadratic formula for an GCSE Maths students looking to stretch themselves a little and go for that A*!
If we have a general quadratic equation:
ax^{2}+bx+c=0 (eq1)
You will have been told that the solution is:
x= (b +/ sqrt[b^24ac])/2a (eq 2)
But why is this true? Let's go through this in steps. Firstly, we will divide our equation by a:
x^{2}+(b/a)x+(c/a) =0 (eq 3)
Now we can use a clever trick called 'completing the square', which states that:
x^{2}+(something)x = (x+(something)/2)^{2}  (something/2)^{2 }(eq 4)
You can check this by multiplying out the brackets on the right hand side!
Anyway, this clever trick tells us that:
x^{2}+(b/a)x+(c/a) = (x+(b/2a))^{2}  (b/2a)^{2} + (c/a) = 0 (eq 5)
Let's tidy up by moving the contants over to the right hand side:
(x+(b/2a))^{2} = (b/2a)^{2}  (c/a) (eq 6)
Now we take the sqaure root of both sides:
(x+(b/2a)) = +/ sqrt[(b/2a)^{2}  (c/a)] (eq 7)
The reason for the +/ is that there are always two answers to the sqrt (for example, (2)^{2 }= 4, but so does (2)^{2}, so sqrt[4] = +/ 2 ). Let's look at the square root on the right hand side and tidy it up a bit:
sqrt[(b/2a)^{2}  c] = sqrt[b^{2}/4a^{2}  c] = sqrt[(b^{2}  4ac)/4a^{2}] (eq 8)
Where in the last step I took out a factor of 1/4a^{2}. We can now use that sqrt[AB] = sqrt[A]sqrt[B] to write:
sqrt[(b^{2}  4ac)/4a^{2}] = sqrt[1/4a^{2}]sqrt[(b^{2}  4ac)] (eq 9)
But 1/4a^{2} is just (1/2a)^{2}, so sqrt[1/4a^{2}] = 1/2a !
So we have:
sqrt[1/4a^{2}]sqrt[(b^{2}  4ac)] = (1/2a)sqrt[(b^{2}  4ac)] (eq 10)
Phew, we're nearly there now! Using this into equation (7) we have:
(x+(b/2a)) = +/ (1/2a)sqrt[(b^{2}  4ac)] (eq 11)
Let's subtract (b/2a) from both sides to make x the subject:
x = (b/2a) +/ (1/2a)sqrt[b^{2}  4ac] (eq 12)
And now (last step!) we take out a factor of (1/2a) and we have at last:
x = (b +/ sqrt[b^24ac])/2a (eq 2)
And we're done!
see moreWhen astronomers discovered Pluto in 1930, they were excited to announce it as the 9th planet in the solar system. The problem is, back then we didn't know that Pluto is just one of many objects floating in a big icy belt we call the Kuiper belt (like the asteroid belt, but much further out).
In order for something to be called a planet, it needs to pass 3 tests:
1. It needs to orbit the Sun (Yep, Pluto does this)
2. Its gravity needs to be strong enough to make it round (Pluto wins here too!)
3. It needs to have 'cleaned up' it's surroundings (oh no!)
The reason Pluto was demoted is because it lives in a really rocky and dusty part of the solar system which it has been unable to clean up.
In fact, we now know of loads of objects of similar sizes to Pluto in the Kuiper belt, so instead of thinking of Pluto not being a planet anymore... you can think of it as the king of the Dwarf Planets!
see more