Need help with this question D:

[man0005]

any ideas?

[Whatlol]

[IMG]http://img109.imageshack.us/img109/7812/physicsqqq.png[/img]



Ok so let the distance from alpha to Y = Ra
Therefore the distance from beta to Y is R - Ra

Now Set the two equations to be equal to each other

Ma/( Ra)^2 =0.01Ma /( R-Ra)^2   ( i omitted  G and m since they will be canceled out)

100/( Ra)^2 =1 /( R-Ra)^2

Ra^2 = 100(R-Ra)^2

edit:
when you get to this point, take the square root of both sides.

Ra + 10Ra = 10R

Ra = 10/11R

[man0005]

thanks! it worked
but yeah i had to use a calculator :/
were you able to do it by hand?

[Whatlol]

ahahha i figured it out, makes it so much more simple

Ra^2 = 100(R-Ra)^2

when you get to this point, take the square root of both sides.

Ra + 10Ra = 10R

Ra = 10/11R

[man0005]

ohhhh LOL
yeah probably would be easier
THANKS

could you please help me with this one? i think heinemann made a mistake with the distances though...

[Aurelian]

ohhhh LOL
yeah probably would be easier
THANKS

could you please help me with this one? i think heinemann made a mistake with the distances though...

Original position, 600,000m (600km) above the surface. Radius of Earth = 6.4*10^6m

Therefore, the original position of the satellite from the centre of the Earth (which is what the graph has) is given as altitude + radius of Earth, ie;

6,400,000m + 600,000m = 7,000,000m = 7.0*10^6m

Final position, we are told, is 8000km from the centre of the Earth (ie 8.0*10^6m). Thus, we find the area under the graph between 7 and 8 by "counting the squares" (lol). It comes to above 7 squares, unless I've counted wrong, which is equivalent to 7.0*10^6J. This is the amount of work done by gravity on the satellite to slow it down from its original speed, and thus is equal to the amount of kinetic energy the satellite has upon being launched.

HOWEVER, that graph, as I notice upon looking at my textbook, is for 1kg, as a consequence we need to multiply our result by the mass of our new satellite (ie 240kg). Thus, 240*7.0*10^6J = 1.9*10^9J

I notice that the answer at the back of the book is 1.7*10^9J. These questions typically have an answer 'range' given the inaccuracy of the "counting squares" method, which I suspect is the reason my answer is slightly off; however, of course, I may have made a legitimate mistake somewhere...

[man0005]

yeah i got 1.9 as well, the counting squares is annoying lol

btw for this question, part d, can you just use the speed you worked out in c to find the answer?
(i know you use the graph, but i just want to check if you should get the same answer if you use g = v^2/r, because i didnt...)

[Aurelian]

yeah i got 1.9 as well, the counting squares is annoying lol

btw for this question, part d, can you just use the speed you worked out in c to find the answer?
(i know you use the graph, but i just want to check if you should get the same answer if you use g = v^2/r, because i didnt...)

Hmm... not sure what way you're thinking of doing it :S

Don't overcomplicate things Just look at the graph and notice that at 2.5*10^6m, the force acting on the rock is 70N. Gravitational field strength is given as N per 1 kg. The Rock is 20kg, and thus the field strength at that point is the force acting on each kg of the rock. Thus, g = 70N/20kg = 3.5Nkg^-1

EDIT: Oh, rereading what you said at the end in brackets there I think you might've known how to do it like that already. K, I get you now. And no, you can't do it that way because it isn't moving in a circular orbit; the object is just heading straight for Earth.

[man0005]

oh yeahh, thanks so much!

[VCEMan94]

can someone please help me with this question?
dont you need q20 to answer q19?
i tried to use the law (T^2 = R^3 ? ) but it didnt work D:

[Aurelian]

can someone please help me with this question?
dont you need q20 to answer q19?
i tried to use the law (T^2 = R^3 ? ) but it didnt work D:

The law isn't T^2 = R^3, it's that for bodies orbiting the same larger body, R^3/T^2 is constant. Thus, you use all the data given to fill in an equation that looks something like R^3/T^2 = R'^3/T'^2, and then solve for the remaining quantity, where R and T are, respectively, the radius and period of the known satellite, and R' and T' are the radius and period of the second semi-known satellite.

[VCEMan94]

ahh yeah
do you remember if your answer for 20 matched the answers?
cause i got 1.5 x 10^22 instead 1.3 x 10^22 kg