As a Theoretical Physics student at Durham University, I am more than aware of all of the confusing turns that science can take. I have a real passion for my subject, and hope to show my students how beautiful science can be.
I am very patient and understanding, and recognise that science can be a very difficult subject to grasp immediately. I have had a considerable amount of experience in one-one tutoring in both Physics and Mathematics, and as such I am aware that everyone learns at different rates. I therefore make sure to tailor every session to the individual student.
Out of all of the subjects taught in schools, I would argue strongly that science is the subject that requires the most understanding. It is not good practice, especially in higher education, to merely memorise facts and figures; one benefits greatly from a confident grasp of the machinery behind the subject.
Thus, I approach every tutorial with a focus on comprehension. There is no point in just showing the student how to solve a problem, as this puts a limit on how many problems they can solve. If the student instead learns not only the how, but also the why of the method, they will be able to use their understanding and intuition to solve many more difficult problems.
Physics and Mathematics are very much subjects where practise makes perfect. I will therefore make certain to familiarise all of my students with the formats of their exams. I am also a firm believer that setting questions which are a little harder than exam questions can do wonders for technique; it is easier to dial back to an easy question than to adapt to a more difficult one.
I am more than happy to answer any questions either via mail or via a Meet The Tutor session.
I look forward to meeting you!
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
|Physics||A Level||£24 /hr|
|Further Mathematics||GCSE||£22 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Julie (Parent) November 23 2016
Okey (Parent) November 13 2016
Okey (Parent) November 6 2016
Julie (Parent) October 26 2016
A body in a stationary orbit will always remain above the same point on the planet as it orbits. For a body to be in such an orbit, it must rotate around the planet in the same direction as the spin of the planet, and its orbital period must be equal to the period of rotation of the planet. In this question we are aksed to calculate the orbital radius at which the satellite will complete one orbital cycle in precisely the time that it will take the planet to complete one full revolution.
The satellite, mass m, will be undergoing uniform circular motion around the centre of mass of the planet at some radius r. The centripetal force, FC, required to keep the satellite moving at a constant angular speed w (where w=2*pi/T; T is the orbital period of the satellite), will be given by
But what gives rise to this centripetal force? Recall that centripetal force is not a force in itself, but rather the name for a force which always acts centrally (towards the same point) on a body undergoing circular motion. In this case, the central force is the gravitational pull of the planet on the satellitle, FG, such that FC=FG. By Newton's universal law of gravitation,
where M is the mass of the planet, and G is the gravitational constant 6.67*10-11 m3kg-1s-2.
Equating the two forces together, we get that
m*r*w2 = (G*M*m)/r2.
We wish to find r, so rearranging to make r the subject and noticing that the mass of the satellite cancels out, we get that
r3 = (G*M)/w2.
We know that w=(2*pi)/T, and we also know that T must be equal to the period of rotation of the planet for a stationary orbit, which we are given. Making this substitution for w, and performing a little algebra,
r3 = (G*M*T2)/4*pi2.
If we substitute in the values of G, M and T, and take the cubed root to get r, we get that
r = 1.3*108 m.
Thus, for our satellite to be in a stationary orbit around this planet it must be 1.3*108 m away from the centre of the planet.see more
Recall that for two lines to be skew they must satisfy two conditions:
1) They must not be parallel.
2) They must not intersect.
We shall check each condition individually.
The general vector equation of a line is given by
r = a + kb,
where a is the position vector of some point on the line, k is a scalar, and b is the direction vector of the line. The direction vector of the line, as the name suggests, dictates in what direction the line travels; b tells us how the line is orientated in space.
For two lines to be parallel, the direction vector of one line must be equal to some scalar multiple of the second line. However, for our two lines, it is clear that there exists no scalar k for which
(-1,2,2) = k(1,3,5).
Thus, the two lines cannot be parallel.
Let us assume that the two lines do in fact intersect. In other words, that
(1,4,1)+s(-1,2,2) = (2,8,2)+t(1,3,5)
for some numbers s and t.
This vector equation leads to three simultaneous equations:
1-s = 2+t (1), 4+2s = 8+3t (2), 1+2s = 2+5t (3).
If we add 2 times Eq. (1) to Eq. (2), we get that
t = -6/5.
If we substitute this value of t into, say, Eq. (3), we get that
s = -5/2.
However, subsituting both of these values into Eq. (2) yields a contradiction. The LHS gives
4+2(-5/2) = 4-5 = -1,
whereas the RHS gives
8+3(-6/5) = 22/5.
Clearly, then, the LHS is not equal to the RHS; the system of equations is inconsistent, and so the lines do not intersect.
We have shown that the given lines satisfy both of the necessary conditions to be classified as skew. The lines are therefore skew, as required.see more
De Moivre's Theorem states that if z = cos(q)+isin(q), then
zn = (cos(q)+isin(q))n = cos(nq)+isin(nq)
z-n = cos(-nq)+isin(-nq).
Now, cos(-p)=cos(p), as cosine is a symmetric (even) function, and sin(-p)=-sin(p), as sine is an anti-symmetric (odd) fuction. Thus,
z-n = cos(nq)-isin(nq).
The rest is just algebra:
zn+z-n = [cos(nq)+isin(nq)]+[cos(nq)-isin(nq)] = 2cos(nq).
zn-z-n = [cos(nq)+isin(nq)]-[cos(nq)-isin(nq)] = 2isin(nq).see more