Looking for a physics expert=)?

Question

I've tried these problems soo many times and my answers are NOT matching up with the answer keys. WIll you do these problems and let me know what you get? I will tell you what I have been getting and maybe you will be able to see what I am doing wrong! Thank you so much for your help!


1. The mass of Earth is about 80 times that of the moon and its radius is about 3.7 times that of the moon. Calculate the acceleration due to gravity at the surface of the moon. Express your answer in g's.
(for this one I got 1.7)


2. Calculate the force of gravity on a 13 kg mass if it were 6.4
*10^6m above Earth's surface (that is, if it were 2 Earth radii from Earth's Center.) The mass of Earth is 6 *10^24

(For this one I got...2.13e29)

3.Calculate the force of gravity between Earth and the sun (Sun's mass = 2.0 *10^30 kg; Earth's mass = 6 *10^24 kg; average Earth-Sun distance = 1.5 *10^11 m).

(For this one I got 0.00592888889)

Answer 1

1- gravity is directly proportional to the mass, and inversely proportional to the SQUARE of the distance. So the earth should have 80 times more gravity than the moon due to the larger mass, and 13.69 times less at the surface (3.7^2) due to the large radius.
80/13.69 is 5.84.
Since we take that gravity on the surface of the earth is 1 g, then on the moon it would be 1/5.84 or 0.17 g. Your answer is off by a factor of 10, so it seems you simply missed the decimal place, a small error that causes a huge variation.


2- For the second problem, you are way off. 13 kg on the surface of the earth weight 127.53 Newton (since the acceleration on the surface is 1 g or 9.81 m/s^2.
Now, you double the distance (2 Earth radii), so the force will be decrease 4 fold (2 squared). So the force on that mass will be 127.53/4 = 31.88 Newton. Notice that the mass of the earth is not involved, unless one insists on calculating the force using G, the universal constant of gravitation, which is 6.67E-11 m^3/(kg s^2)

3- For the 3rd one, one needs the big equation

F = G m1 m2 / r^2

And the force is huge: 3.5E22 Newton (3.5 * 10^22)

This is what is needed to keep the earth in place, as it travels at 30 km/second in its orbit.


You have to keep better control of what you multiply and what you divide. Seems to me you tend to confuse those.

Answer 2

I won't work the arithmetic, but I will explain the physics, which is really what you should be learning anyway.

F = GMm/R^2; is old Newt's gravity law. F = mg is one of his three laws of motion in terms of g as the acceleration due to gravity on Earth = 9.81 m/sec^2. m is the body gravity is acting on, M is the body with the pull, G is a constant of proportionality, and R is the radial distance of m from the center of M.

Let's say F relates to the force of gravity on Earth's surface and f = G(M/80)m/r^2 and f = ma relates to that same force, but on the surface of the moon of radius r and mass (M/80). a = the gravitational acceleration on the moon and the answer you are looking for in terms of g, which is the acceleration due to gravity on Earth.

1) Whenever you are looking for an answer in terms of another factor (like a in terms of g), start with a ratio, like f/F = G(M/80)m/r^2//GMm/R^2 = (1/80)(R^2/r^2); so that f = ma = mg(1/80)(R^2/r^2) and a = g(1/80)(R^2/r^2); where R = 3.7r and you have enough to work the problem and get a as some multiple of g.

2) Again, invoke Newt's law of gravity and do a ratio with F = the force at the Earth's surface and f = the force at 2R. Thus we have f/F = GMm/(2R)^2//GMm/R^2; so that f = F(1/4) =mg/4; where m = 13 kg and g = 9.81 m/sec^2 You now have enough to work the problem. (You also have an important physics law, the force of gravity varies as the square of the distance between the two attracting bodies.)

3) F = GM(M/.3 X 10^6)/R^2; where M is the mass of the Sun, (M/.3 X 10^6) is the mass of Earth "in terms of Sun mass", G is the gravitational constant, and R is the distance between the mass centers of the Sun and Earth. If you don't know G, you can solve for it using your answers in 2). f = GEm/(2R)^2; where f is the force on the 13 kg mass (m) that you found earlier. So G = f(4R^2)/Em and all the factors on the RHS are known; where E = Earth's mass, m = 13 kg, and R = Earth's radius.

Lesson learned: When finding something in terms of something else (like a in terms of g), start with a ratio (like the f/F). You will likely find that many factors cancel out (like the M, m, and G).

Answer 3

1)
F = (G.M.m) / r²
G is the universal gravitation constant,
M is the mass of earth or moon ( considering the both instances) ,
m is the mass of the object considered in both instances
r is the distance to the object from earth or moon ( considering both instances )

G and m are constant in both instances

so F α M / r²

consider acceleration due to gravity in the moon is g'
and acceleration due to gravity in the earth is g
moon's mass is M' and earth's mass is M
moon's radius is R' and earth's radius is R

F is the gravitation force in other words weight
d is the the distance of the object from the surface
(we'll consider on both instances the distance from the surface is same)

in the earth the considered object's weight = mg
in the moon considered object's weight = mg'

so in the earth
mg α M / (R + d)² ---------(1)

in the moon
mg' α M' / (R' + d)² ---------(2)

but because the object is on the surface and very close to surface
R >> d
so (R + d)² ≈ R²

in the same way
(R' + d)² ≈ R'²

so (1) and (2) are re-written as

mg α M / R ² ---------(3) and
mg' α M' / R' ² --------(4)

to get rid of the proportional mark

(3) / (4)
g / g' = MR'² / M'R²

g' = g M'R² / MR'²

g' = g [ M'(3.7R')² / [ ( 80M')R'² ]
g' = (3.7)² / 80g
g' = 13.7 / 80g
g' = 0.17g

that means gravitation acceleration on moon is 0.17 times gravitational acceleration on earth.

2) F = G.M.m / r²

earth's radius = 6.4 * 10^6 m when considering data
F = (6.69 * 10^-11)*(6 * 10^24 )*(13) / (2 * 6.4 * 10^6 )²
F = [6.69 * 6 * 13 / (12.8)²] * 10^1
F = 5218.2 / 163.84
F = 31.85 N

3) F = G.M.m / r²
F = (6.69 * 10^-11)*(2 * 10^30)*(6 * 10^24 ) / (1.5* 10^11 )²
F = (6.69 * 2 * 6 /2.25 ) * 10^21
F = 35.68 * 10^21 N

Answer 4

You seem to be way off on #2 and 3. Did you use F = GMm/r^2, where G = 6.67 x 10^(-11), M and m are the masses of the two objects, and r is the distance from the center of one to the center of the other?

Also, you seem to be having difficulty with exponents. Examples:

10^30 times 10^24 = 10^54. (Add the exponents.)
10^30divided by 10^24 =10^6 (Subtract exponents.)
(10^6)^2 = 10^12 (Multiply the exponents.)

Answer 5

1. 9.8 x (3.7^2 for smaller radius) / 80(for larger mass)

2. weight on earth = mg = 13*9.8. Divide by 2^2 if you are twice as far from the center

3. F = Gm1m2/d^2 Use the exp button on your calculator and don't forget tp square d

Answer 6

well i can't do it with out gravtiational constant, but 1) is g=Gm/R^2
and just use earth mass and radius but, divide by 80, and the radius by 3.7
2) F=GmM/R^2
3)same formula