1,000,000 Question Thread :D

[Akirus]

You need to use the law of conservation of energy, in this case.

The initial mechanical energy when it is thrown (kinetic energy + gravitational potential):






Then we calculate the gravitational potential energy when it is 10.0m from the base of the cliff and subtract it from this total (since no energy is lost, only converted; as the rock falls, it loses and gains ):





Now subtract it:







This figure can also be used for other things, such as finding the velocity of the stone when it is 10.0m from the base of the cliff:







This can be verified with Newton's equations (another way to do this question, as well):















This proves that the velocity is the same no matter how you work it out. And now that we have the velocity, we can also solve for :

[kenhung123]

Got a few more physics questions: 7b, 8 and 12

[/0]

sup kenhung,

7b) If



Let the distance from Alpha be 'd', then:



Then solve for d.

8. Let the distance from Alpha be 'x', then:





Then solve for x.

12. Let the distance from Earth be

Then



Then solve for

[kenhung123]

Thanks a lot. Umm, but I don't really get how to figure out which is (r-d) and d for different formulae and how do you solve that by scientific calc?

[/0]

In 7b) it says "what distance... from the planet Alpha", so since that is the unknown, you give a name it, for example 'd'.
Similarly in 8 and 10 you have "distance from Alpha" and "distance from centre of Eath" respectively.
In all cases, since the total separation is , and the distance from one planet to your point is , the other distance must geometrically be .


Assuming you don't have a 'solve' button on your scientific calculator, you'll just have to do it by hand

In 7b),





(cross multiplying)





(or quadratic formula)

(d > 0)

The others are similar

[moekamo]

(cross multiplying)

instead of expanding from here, another way is to take the square root:

(cross multiplying)



(square rooting)

[kenhung123]

Thanks guys for all the help

[kenhung123]

Ok 10d is a bit strange. I used v^2/R but didn't work.

[/0]

At m from the centre of Mercury, the force on the 20kg rock (from the graph) is .

So the gravitational field strength is

[kenhung123]

Oh right, but how come V^2/R doesn't work?

Also this question 12, R^3/T^2=GM/4pi^2 doesn't work too.

These questions are confusing as some formula don't work

[/0]

would work if the rock were in circular orbit. However, it's speeding towards mercury, so the formula doesn't apply.

However, Kepler's 3rd Law should work for this question. Make sure you convert years into seconds.

[kenhung123]

Yep, I have, still doesn't work. The answer seem to have used kepler's law with an extra small 'm' which I don't know why

[/0]

hmmm, well...



and IS the mass of the sun.

So whatever the answers say, if you got something close to that it should be right.

[kenhung123]

Lol dam! I forgot the squares and cubes in calculator!

Anyway, could you confirm that the exceptions to the gravitation formulae? So all the formulae only apply to objects in a stationary orbit? Kepler, v=2piR/T, a=v^2/R, F=v^2m/R, F=4pi^2R/T^2 etc etc

[/0]

, , , are all for objects in circular orbit.