I've understood Physics up to this point, any help is greatly appreciated. Even if its just an answer to one question!
Thank you!
1.
a. A small light source located 1 m in front of a 1 m2 opening illuminates a wall behind. If the wall is 1 m behind the opening (2 m from the light source), the illuminated area covers 4 m2. How many square meters will be illuminated if the wall is 3 m from the light source?
b. What if the wall is 5 m from the light source?
c. What if the wall is 9 m from the light source?
2. Calculate the force of gravity on a 8 kg mass at Earth's surface. The mass of Earth is 6. 1024 kg, and its radius is 6.4 106 m.
3. Calculate the force of gravity on a 10 kg mass if it were 6.4 106 m above Earth's surface (that is, if it were 2 Earth radii from Earth's center.) The mass of Earth is 6. 1024 kg.
4. Calculate the force of gravity between Earth (mass = 6.0 1024 kg) and the moon (mass = 7.4 1022 kg). The average Earth-moon distance is 3.8 108 m.
5. Calculate the force of gravity between Earth and the sun (Sun's mass = 2.0 1030 kg; Earth's mass = 6. 1024 kg; average Earth-Sun distance = 1.5 1011 m).
6. What would be the difference in your weight if you were six times farther from the center of Earth than you are now?
Weightnew / Weightcurrent =
b. What would the difference be if you were thirteen times as far?
Weightnew / Weightcurrent =
Answers & Comments
Verified answer
1. Since at 1m from the source the source goes through a 1m^2 gap, and at 2m the source is 4m^2, you've got the square of the distance (that can be mathematically proven using angles and stuff).
a. Therefore at 3m, 3^2 = 9m^2 of wall will be illuminated.
b. 5m, 5^2 = 25m^2 of wall illuminated.
c. 9m, 9^2 = 81m^2 of wall illuminated (it'd have to be a strong light though!)
2. All you need here is the formula for the force of gravity: Fg = GMem/Re^2 (G = universal gravitational constant, Me = mass of the earth, m = mass of the object, Re = radius of the earth)
Since G = 6.67*10^-11,
Fg = 6.67*10^-11 * 6*10^24 * 8 / (6.4*10^6)^2
Fg = 78.16N
(A good check which works for cases on the surface of the earth is to multiply m by 9.81 as we know this is the size of the acceleration of gravity: 8 * 9.81 = 78.48, so all good!)
3. is just the same, but now m -> 10 and Re -> (6.4*10^6) * 2, the best way to learn is by doing yourself :)
4. This one is quite cool, havn't seen a question like this before.
So you've still got your formula for Fg. G and Me are still going to be the gravitational constant and the mass of the earth, but now m will be the mass of the moon (7.4*10^22), and R will now be the distance from the earth to the moon (3.8*10^8). (I will be referring to Re as R now, as we are no longer talking about the radius of the earth.)
5. Exactly the same as the previous one, but now with the suns mass and distance away from the earth instead of the moons :)
6. a. At the moment, you are Re away from the earth's core. If you increase this to 6Re, remembering our formula uses Fg = GMem/Re^2, so Re has become 6Re, therefore GMem/(6Re)^2 and therefore we have a factor of 36 on our denominator now. Therefore, we are 36 times as light! Or 1/36th of our weight, when only 6 times further away from the earth's core.
b. Same logic, but now we have a factor of 13Re. Our denominator becomes (13Re)^2, or 169. So we are 169th of our weight when 13x further away from the earths core than currently.
(Adding on to that, when only increasing our distance from 6x Re away to 13x Re, just over double, we get almost 8x as light. :) )
Hope this helped and you weren't just looking for some quick answers!
2) 6.67384 * 10^-11 * 8 * 6.0*10^24 / (6.4*10^6)²
= 78.2 N
5) 6.67384 * 10^-11 * 2.0*10^30 * 6.0*10^24 / (1.5*10^11)²
6.67384 * 2.0 * 6.0 * 10^43 / (2.25 * 10^22)
= 3.559 * 10^22 N
6) a) 1/36
b) 1/169
Find help for your Gravity Problems here:
http://www.physicsclassroom.com/class/circles/u6l4...