could you please lend me a hand with some problems?

[run-bandit]

1) venus and saturn both orbit the sun. using only ifnromation about the sun and the periods of the two planets,m calculatoe the value of the ratio of :


 distance of saturnfrom the sun/
distance of venus from the sun


2) explain why the area under a gravitation force-distance graph gives the energy needed to laungch a satellite but the area under a graviational field stregnth-distance graph gives the energy per kilogram needed to launch a satellite?

thank you!

[appianway]

1) T(saturn)^2/T(venus)^2 = R(saturn)^3/(R(venus) ^3 using kepler's laws.
Thus the ratio of the distances is the cube root of T(saturn)^2/T(venus)^2

2) E = (integral ) F ds. Integrating gives the area under the graph, and hence by placing the limits on the energy at the earth's radius and the final height, you'll get the energy needed to launch. For the field graph, E = (integral) mg ds, as mg gives the force (as it does in the first one too). As m is constant, it can be brought out of the integral, and if both sides are divided by m, (integral) g ds = E/m.

[run-bandit]

:s

I don't quite follow what you are saying

[kamil9876]

As for the area under the graph stuff, this was sort of explained here.

Note: calculus is not necessary for vce physycs, but the ideas of calculus are implicitly used in VCE physics and so it makes stuff easier if you at least sort of understand the basic conceptual idea without neccesarily knowing the technical manipulations etc.

edit: I'll paraphrase my post from that thread which I think applies here:

You know that work done is Hence if the force is not constant you have to split the interval up into many pieces. Basically, the work done by moving the object from 1m to 2m is equal to the work done by moving it from 1m to 1.1 plus work done from 1.1 to 1.2. etc. And for that interval of 0.1m u can assume the force is constant (the numbers are an unrealistic example, but i hope u get the idea). Basically you want to take the limit as the interval approaches zero, and you can see that this is analogous to finding an integral [area under the graph].