apparent weight physics question?

Question

apparent weight physics question?

Answer 1

Small correction on that formula. It's r^2, not d^2. The correct formula is Fg = G[m1m2] / r^2



G = gravitational constant (6.673E-11)

m1 = mass of object one in kg (the Moon)

m2 = mass of object two in kg (the astronaut)

r = distance from the center-to-center of the two objects in meters



Also, you need to find the mass of the astronaut, because it appears you only have the astronauts weight on Earth. Use the formula m = W/g



m = mass of person

W = weight of person

g = gravity of Earth (9.806 m/s^2)

Answer 2

1) By F(net) = mg + mv^2/r =>F(n) = 90 x 9.8 + [90 x (960 x 1000/3600)^2]/900 =>F(n) = 7993.11 Newton 2) As W' = 7993.11 ≈ 9g =>If blood is allowed to pool in the lower areas of the body, the brain will be deprived of blood, leading to temporary hypoxia. Hypoxia first causes a greyout (a dimming of the vision), also called brownout, followed by tunnel-vision and ultimately complete loss of vision 'blackout' followed by g-induced Loss Of Consciousness or 'G-LOC'. The danger of G-LOC to aircraft pilots is magnified because on relaxation of g there is a period of disorientation before full sensation is re-gained. A G-suit does not so much increase the g-threshold, but makes it possible to sustain high G longer without excessive physical fatigue.

Answer 3

How is the space vehicle moving at a constant velocity? Is it in an orbit around the moon? In that case too, there would be acceleration. If the spacecraft were falling freely towards moon, the apparent weight would be zero. So, take a fresh look at the problem.

Answer 4

that's the correct formula

but

you have to find the F with respect to Earth as well as F with respect to the Moon

then

the weight ( net F ) is the vector sum of these two

(one is positive and one is negative - opposite directions - so it will be the difference between them)

Answer 5

Answer --> Total apparent weight is 40.61 N, toward the Moon.



I could be wrong, but I think that everyone's making this much more difficult than it has to be. At the same time, they're making it far too simplistic. Assuming that the Earth, astronaut, and Moon are momentarily all in perfect alignment.



Given:



m = 70 kg

r = 2,900,000 m



Weight is a force, mass multiplied by local gravity reference.



W = m * g



You have mass, now you just need to find gravitational acceleration at that distance. This is found by:



a = GM / r^2



Where:



G = Universal Gravitational Constant

M = Mass of the Moon



Insert UGC, look up M:



G = 6.67428E-11 m^3/kg-s^2

M = 7.3477E+22 kg



Solve for gravitational acceleration at that altitude.



a = [ (6.67428E-11 m^3/kg-s^2) * (7.3477E+22 kg) ] / (2,900,000 m)^2

a = [ 4.904E+12 m^3/s^2 ] / (8.41E+12 m^2)

a = 0.583 m/s^2



Find simple weight



W = (70 kg) * (0.583 m/s^2)

W = 40.8 N



If you want to get more technical, take this into account. The center-to-center distance between Earth and Moon is 384,405,000 m. So taking away 2,900,000 m, leaves you with a distance from the center of the *Earth* as being 381,505,000 m. Look up Earth mass:



M = 5.9736E+24 kg



Resolve gravitational pull from Earth at that distance.



a = [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (381,505,000 m)^2

a = [ 3.987E+14 m^3/s^2 ] / (1.455E+17 m^2)

a = 0.00274 m/s^2



So the apparent weight *from* Earth is



W = (70 kg) * (0.00274 m/s^2)

W = 0.19 N



If you subtract the Earth weight from Moon weight (they're acting against each other), you get the astronaut's apparent weight:



Wa = (40.8 N) - (0.19 N) = 40.61 N, toward the Moon.



Keep in mind that "toward the Moon" is very important, because a force is a vector quantity, with both a magnitude and a direction.