Recent content by learningisfun

oh thanks for moving the post, I will look around here

HI I'm not to sure where to place this post. I'm looking for a computer application where I could plug values into get charts , bar graphs pie charts etc. I do not have very much experience in this field , but I want to play around. If you guys know of any great programs that can produce...

hmm, ok that is interesting to read. This does not answer my question? actaully hold on, I"m working on it

ok The neutral point between the moon and the earth m(moon)7.35E22 m(earth)5.98E24 R(between Earth and moon)3.84E5km=3.84E8m G6.67E-11v I used qudratics this time , now I'm stuck and will tackle it again after i Get some help ^.^ GMmM/r^2=GMeM/(L-r)^2 Mm/r^2=Me/(L-r)^2...

I remember in my science class back , my teacher would say stuff like , star wars has very little true facts about science, their is no sound of explosion ins space. THen amake a relevant connection to todays pop culture like Star Wars. XD

Find the distance when the opposing forces of the Earth and the moon become neutral, relative to earth. I did it in the above post, but I would like to do it differently using quadratic equations or substitution?

apparently you could use quadratic equations, can someone explain how to do that

This time I worked out the gravaitonal forces of the Earth and the moon. By dividing it I can gather a percentage of the difference in magnitude. Fgearth/Fgmoon=% G=6.67x10^-11 m1=5.98x10^24 m2=7.35x10^22 r1=6.38x10^6 r2=1.74x10^6 Fgearth=Gm1m/r1 =(6.67x10^-11)(5.98x10^24)(1)/6.38x10^6 =9.8N...

hmm, I have another equation in mind, take a moment to do it

I'm trying to find out the gravitational point of the moon and Earth it's one of my questions -.- I did not realize that I forgot to include that in , ehm sorryFg=Gm1m2/r^2 =6.67x10^-11(5.98x10^24)(7.35x10^22)/(3.84x10^8)^2 =3.41x10^30 I'm not sure waht to do with Fg next

because Fg=Gm1m2/r^2= mv^2/r thierfore r=sqaurerootGm1m2/Fg ok I got 2.93x10^14 this is murch greater then the distance between the moon and Earth at the center, I need to approach this differently. Somebody help me out >.> hold on I have another fourmla in mind, to get r r=Gm/v^2 in that...

wait what is the graviational force between both spheres m1=5.98x10^24 m2=7.35x10^22 r=3.84x10^8 Fg=Gm1m2/r^2 =6.67x10^-11(5.98x10^24)(7.35x10^22)/(3.84x10^8)^2 =3.41x10^30 the gravitational force between hte moon and the Earth is =3.41x10^30 (check my answer please to make sure I'm...

oh , that's amazing, the sun is far larger, *draws out sketch* troginnometry ^.^

well objects in uniform motion experience a force acting to accelerate the object towards the centere of curvature. Finding the Fg between the center of Earth and 1 kg on hte surface being 9.8N. Is their a cut of radius where the mass recedes from center?

Another noob qeustion that I do not have the knowledge to figure out. It's is possible to see the shadow of a sattilte?