User Controls
A trianglism math question for those who are worthy.
-
2016-09-11 at 9:05 PM UTC
If it's 9,807 at 0 distance then how can I calculate that at 384.4km?. Twice as far away from 0 distance is 0 distance. If you said 9,807 at 10km distance, i would've been like 9,807/4 = 20km, 9,807/16 = 40km, 9,807/64 = 80km.
I already gave you a baseline idiot. 9,807m/s2. Here is the formula. Fgrav = G * M☉ * m over d2. Here's a visual aid.
Also the dot here, is not a decimal so you can't round to 384.4km, it's threehundredfortyhousand kilometres. You're solving for the strength of the force. In the first equation above, g is referred to as the acceleration of gravity. Its value is 9,807m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2 rounded. M☉ is the mass of the earth and you can cancel for the small m. Where d represents the distance from the center of the object to the centre of the earth. The dist6ance from earth's centre at sea level is 6.38 x 106 metres. -
2016-09-11 at 9:26 PM UTCRealistically would you be able to deduce newtons law of gravitation without any prior knowledge and from your question alone?
-
2016-09-11 at 9:56 PM UTC
I already gave you a baseline idiot. 9,807m/s2. Here is the formula. Fgrav = G * M☉ * m over d2. Here's a visual aid.
Also the dot here, is not a decimal so you can't round to 384.4km, it's threehundredfortyhousand kilometres. You're solving for the strength of the force. In the first equation above, g is referred to as the acceleration of gravity. Its value is 9,807m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2 rounded. M☉ is the mass of the earth and you can cancel for the small m. Where d represents the distance from the center of the object to the centre of the earth. The dist6ance from earth's centre at sea level is 6.38 x 106 metres.
"F sub g is the gravitational force, m1 and m2 are the masses of the two objects, r is the separation between the two objects and G is the Universal Gravitational constant. This is how the force of gravity between two objects is calculated."
Wouldn't that mean 9,607 is the gravitational constant, but you're saying that's the acceleration on earth? That doesn't sound right, if it's only "on earth" then it's not a constant. You don't cancel for small m when you're using the mass of the moon though. So you canceled for small m because it's the mass of earth. the second time, how do you know just because it's redundant you're supposed to remove it from the formula completely? And if you're calculating the gravitational force of earth by the distance of the center to the sea level and that produces a different result than the gravitational acceleration (which is supposed to be a constant) on earth, how are the two things different? You're talking about calculating the force of gravity on earth from the acceleration of gravity on earth (when it's actually supposed to be a gravitational constant) and it produces a different result from the the original force? I don't really have any clue wtf you're talking about but are you sure you do? [(9,807m/s2)gravity acceleration x (5.972 × 10^24 kg)mass of earth x 7.34767309 × 1022 kg)mass of moon]/384.400km^2. WAT. How do I even add m/s2 to kg when they're completely different units, and do I have to actually square the 384.400km? If I do all of that, ignoring the units and just adding and diving everything
[9.807+(5.972 × 10^24)+ (7.34767309 × 10^22)] / 384.400^2 =
wtf is this wtf am i doing -
2016-09-11 at 10:19 PM UTCI like your graph. And i also like how much effort you put into this, there may be hope for you yet. You should enroll into college and show everyone that you're actually a smart guy. The trick here was that you need to solve for a force, but the contextual clue was the strength of gravity in acceleration. Calculate force convert to the strength of gravity in acceleration. Being in an orbit is analogus to falling constantly but having enough velocity to keep missing the the object you're orbiting.
-
2016-09-11 at 10:46 PM UTCthis is my answer
-
2016-09-12 at 3:56 AM UTC
I like your graph. And i also like how much effort you put into this, there may be hope for you yet. You should enroll into college and show everyone that you're actually a smart guy. The trick here was that you need to solve for a force, but the contextual clue was the strength of gravity in acceleration. Calculate force convert to the strength of gravity in acceleration. Being in an orbit is analogus to falling constantly but having enough velocity to keep missing the the object you're orbiting.
correct answer
6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 = f
my answer
[9.807+(5.972 × 10^24)+ (7.34767309 × 10^22)] / 384.400^2 = f
So actually I was right, but the fact that you masked the Universal Gravitational Constant in an unnecessary figure of earth's gravity acceleration tricked me and triggered me, causing me to have the wrong value for the G variable when I could've just googled the number. I don't know why it came out as a graph and not a number though, must be the formatting. -
2016-09-12 at 6:03 AM UTCOnly is space do shitty pseudomath threads turn into legit discussions of physics.
-
2016-09-12 at 8:31 AM UTCKek, I was going to come into this thread to explain how participation in mathematics is contingent upon, if not reducible to, the fairly constrained social elements of the practicing community (a narrow set of notations, common language, standards of rigor to say nothing of american pragmatist views of the field) but then I saw advertising to random masturbating adolescents on omegle and decided that was sufficient to redeem any prior transgressions against mathematics.
-
2016-09-12 at 1:05 PM UTC
correct answer
6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 = f
my answer
[9.807+(5.972 × 10^24)+ (7.34767309 × 10^22)] / 384.400^2 = f
So actually I was right, but the fact that you masked the Universal Gravitational Constant in an unnecessary figure of earth's gravity acceleration tricked me and triggered me, causing me to have the wrong value for the G variable when I could've just googled the number. I don't know why it came out as a graph and not a number though, must be the formatting.
Well, plugging the proper numbers into the equation is not solving the equation, but it's a step in the right direction. It's setting up the problem in order to solve it, you still need to give me the strength of the force and convert it. Also, part of the problem in my thinking was inferring the proper formula to use. -
2016-09-12 at 3:53 PM UTC
participation in mathematics is contingent upon, if not reducible to, the fairly constrained social elements of the practicing community (a narrow set of notations, common language, standards of rigor to say nothing of american pragmatist views of the field)
Exactly, which is why trianglism > sciensism. If you can fold a piece of paper and not be a plebeian you can figure it out.
