kk got it thanks. Hey can I ask you guys another question? What exactly is the difference between centripetal force and gravitational force?? Are they the same? WHY IS acceleration (of free fall) the (resultant) force of the difference between gravitational and centripetal force??
Why do attractive force have a negative sign? The book says its because one must be moving toward the other. But how does it make it negative? I dont understand. Can anyone explain the law of gravitation formula?? F= -GMm/r^2
Yes yes, that substitution escaped me......I did it that way now. But i am not getting the same answer as i did in the mgh way...any suggestions on why?
I also know mar's gravitational acceleration. But like i said using GMm/r cannot yield anything because like you know, the masses are in ratio. So I dont know the mass! How can I proceed then??
Are you sure your method will works? u say ΔKE+ΔPE=0 where ΔPE=(GMm/R).(R/R). But you can't use the formula ΔPE=(GMm/R).(R/R) in this way, you don't know the mass of mars!
Homework Statement
The mean diameters of Mars and Earth are 6.9 10^3 km and 1.3 10^4 km, respectively.?
The mass of Mars is 0.11 times Earth's mass.
(a) What is the ratio of the mean density of Mars to that of Earth?
(b) What is the value of the gravitational acceleration on Mars...
he full acceleration due to gravity with its appropriate sign would be -g. So why are we not using the this? If it were then we would get (-W), and if we plug this value into the equation
N-W= ma we would get N-(-W)=ma which would become N+W=ma. My method is to first find why the elevator is...
You are saying W=(-9.81m). Right? So now If i plug it into the equatio of N-W=ma won't I end up with N+W=ma ?? And maybe I know this already, but what exactly is the force law?
(Before I start, due to the use of vectors let me state I consider up to be positive and down to be negative)
Imagine a scenario, of a man in an elevator. And the elevator is accelerating upwards. Now in this case the Normal force > weight. So Normal Force - Weight =mass X acceleration. If...
Well its just a rough variable, think of you driving in a constant circular motion. Once in a flat road and then in a banked road. If you were to keep all variables the same, velocity, road surface, tyre grip etc which terrain would you find easier to maintain the circular motion?