Calculating the mass of the moon given radius.

[cruisx]

Homework Statement

Hi guys so i have this problem that i have been trying to solve and i think i might have the right answer but i am a bit confused about something.



An astronaut conducting experiments on the moon, drops an object from a height of 1.6m and notices that it takes the object 1.4 to reach the moons surface. If the mean radius of the moon is 1.74 * 10⁶m , calculate the mass of the moon.

Homework Equations

gx = GMx/Rx²

Mx = gx * R² / G

The Attempt at a Solution

Well this is where i get confused, I know that the gravity on the moon is 1/6th that of Earth so its equal to 1.6m/s2.

So this is what i did

Mx = 1.6m/s2 * ( 1.74 * 10⁶m )² / 6.67 * 10^-11

and i got the mass to be 7.26260869 * 10²²

Is this the correct answer?

 

[cepheid]

Well this is where i get confused, I know that the gravity on the moon is 1/6th that of Earth so its equal to 1.6m/s2.

Rather than using your existing knowledge about the moon to figure out what g is, I think that you are supposed to use the kinematics data given in the problem (how long it took an object to fall freely from a given height) to *calculate* g. Then you can proceed given the knowledge that g = GM/R²

 

[cruisx]

Ok so i think this is rigth but i don't know. So i used the 3rd motion eqn to find a.

d = T +1/2a(T)^2

and then i found acceleration to be -2.1952 m/s^2

IS that the correct value i should be using for g?

 

[cepheid]

IS that the correct value i should be using for g?

No, because the formula is actually:

d = v₀t + (1/2)at²​

where v₀ is the initial velocity. In this case, the initial velocity of the dropped object is zero, so the formula reduces to:

d = (1/2)at²​

Using this formula will give you an answer close to the 1.6 m/s² that you remembered.

 

[cruisx]

No, because the formula is actually:

d = v₀t + (1/2)at²​

where v₀ is the initial velocity. In this case, the initial velocity of the dropped object is zero, so the formula reduces to:

d = (1/2)at²​

Using this formula will give you an answer close to the 1.6 m/s² that you remembered.

Oh yes, that was a silly mistake, i forgot about that. And you so after your suggestion i got the value fo g to be -1.568 m/s2

And then i pluged it into the formula i mentioned in my first post and got the mass to be

-7.117356522 * 10²²

 

[D H]

Well this is where i get confused, I know that the gravity on the moon is 1/6th that of Earth so its equal to 1.6m/s2.

So this is what i did

Mx = 1.6m/s2 * ( 1.74 * 10⁶m )² / 6.67 * 10^-11

and i got the mass to be 7.26260869 * 10²²

Is this the correct answer?

That is not the correct answer, correct in this case meaning the answer the instructor expects you to get using the given information. In particular, you did not use the experimental evidence.

 

[D H]

Oh yes, that was a silly mistake, i forgot about that. And you so after your suggestion i got the value fo g to be -1.568 m/s2

And then i pluged it into the formula i mentioned in my first post and got the mass to be

-7.117356522 * 10²²

Does that answer make any sense? First off, no units. 7*10²² what? Secondly, when is mass ever negative?

 

[joshmdmd]

I have been assigned this question..
the answer according to my worksheet is given as 7.4E22kg

I cannot get this answer by finding a through a=d/t^2,

My attempt:

Fnet=ma=Fg=Gmm/r^2
ma=Gmm/r^2
a=Gm/r^2
(r^2*a)/G = 3.631304348E22 (its half of the actual answer.. what am I doing wrong??)

 

[cepheid]

I cannot get this answer by finding a through a=d/t^2,

Look at the posts above. From the kinematics equations for constant acceleration, it should be a = 2d/t². Perhaps that is the cause of the problem you are having?