Does this question make sense?

toastedpickles

Hi guys, I'm a bit out of my area here but I'll ask anyways, this is the question

Measure the gravitational force of planet earth on a sphere

we can pick any size for the sphere.
Does this make any sense to you guys? My friend rang his brother, who is a physicist and even he said it made no sense

Thanks!

CramCycle

toastedpickles wrote: »

Hi guys, I'm a bit out of my area here but I'll ask anyways, this is the question

Measure the gravitational force of planet earth on a sphere

we can pick any size for the sphere.
Does this make any sense to you guys? My friend rang his brother, who is a physicist and even he said it made no sense

Thanks!

The size does not matter but you would need the mass AFAIK.

toastedpickles

CramCycle wrote: »

The size does not matter but you would need the mass AFAIK.

Riiight, but how do you go about doing that? say we have a sphere of 1000 kg

We have a very vague lecturer so we have to kind of make our own stuff as we go along

Is it just 9.81 by the spheres weight?

sebastianlieken

yep the question makes sense. kind of. (maybe it's a trick question that was posed to you)

"on a sphere" is silly wording. It doesn't matter if an object is a sphere or a pyrimate or a cat. what matters is the mass of the object; in this case the sphere.

It's Newtons law of universal gravitation. Basically, every coupling of objects in the universe is attracted to eachother, and that attraction force is directly proportional to the mass of both objects, and inversly proportional to the to the square of the distance between them.

the law is very simple,



if you take you, M1 = 70 Kg
and take planer earth, M2 as 5.972x10^24 Kg

r is the distance between your centre of mass and the centre of the earth. say, 6,371,000 meters. (ignore your height as it's negligable)

G = the gravitational constant = 6.67384 × 10-11 m3 kg-1 s-2

solve the equation for "F" yourself. it's my lunch time :P

But I shall leave you with this little tid-bit. Fat people ARE actually more attracted to eachother. :P


Now that i've written all that, it just dawned on me that all your lecturer really cares is F = ma. lol. ("a" in this case is a constant vector (on earth)...and is 9.81 m/s^2, and "m" is the mass of your shere)

toastedpickles

sebastianlieken wrote: »

yep the question makes sense. kind of. (maybe it's a trick question that was posed to you)

"on a sphere" is silly wording. It doesn't matter if an object is a sphere or a pyrimate or a cat. what matters is the mass of the object; in this case the sphere.

It's Newtons law of universal gravitation. Basically, every coupling of objects in the universe is attracted to eachother, and that attraction force is directly proportional to the mass of both objects, and inversly proportional to the to the square of the distance between them.

the law is very simple,



if you take you, M1 = 70 Kg
and take planer earth, M2 as 5.972x10^24 Kg

r is the distance between your centre of mass and the centre of the earth. say, 6,371,000 meters. (ignore your height as it's negligable)

G = the gravitational constant = 9.81 m/s^2

solve the equation yourself. it's my lunch time :P

We've been trying to solve this all morning! :P Thanks all the same though it's more help than we got so far

zl1whqvjs75cdy

Just pick a sphere the size of the earth. Job done

sebastianlieken

toastedpickles wrote: »

We've been trying to solve this all morning! :P Thanks all the same though it's more help than we got so far

I mistyped the gravitational constant in my post. fixed now. (changed something else aswell at the bottom)

everything that you need is in my post.

just a formula to plug some numbers into now.

shout back if you need more.

toastedpickles

sebastianlieken wrote: »

I mistyped the gravitational constant in my post. fixed now. (changed something else aswell at the bottom)

everything that you need is in my post.

just a formula to plug some numbers into now.

shout back if you need more.

Thanks captain

Carawaystick

Is this related to Gauss Law?

If the sphere is hollow ( a shell) and outside the earth

toastedpickles

Carawaystick wrote: »

Is this related to Gauss Law?

If the sphere is hollow ( a shell) and outside the earth

Nope, we think he means it something like the earth and the moon, that kind of concept

sebastianlieken

toastedpickles wrote: »

Nope, we think he means it something like the earth and the moon, that kind of concept

the answers are (for the moon and earth),

F_perigee = 2.21884x10^23 N (N is shorthand for Newtons, and are your units)
F_apogee = 1.78097x10^23 N

the moon follows an elliptical orbit around earth. It's closest at the 'perigee', and furtherst at the 'apogee'. (apogee/away).

So you would expect the Force to be weaker when the moon is at it's furthest from the earth... which it is.

BAM! SCIENCE!

toastedpickles

sebastianlieken wrote: »

the answers are (for the moon and earth),

F_perigee = 2.21884x10^23
F_apogee = 1.78097x10^23

the moon follows an elliptical orbit around earth. It's closest at the 'perigee', and furtherst at the 'apogee'. (apogee/away).

So you would expect the Force to be weaker when the moon is at it's furthest from the earth... which it is.

BAM! SCIENCE!

We still can't get it, the joys or structural mechanics eh? apparently his "logic" is that the moon and earth rotate together instead of the moon around the earth, We've given up

Although i learned more from you guys in 10 minutes than i have from this guy all semester

sebastianlieken

the moon orbits the earth, the earth (and moon) orbit the sun, the sun is spinning in the milkey way, and the milkey way is wizzing through the universe, etc etc... that doesn't matter one bit when it comes to gravitational forces. A gravitational force is an attraction between 2 objects as far as you're concerned - when you start factoring in every other factor you're delving into quantum mechanics which is lunacy if you havn't even been thought the basics yet.

Yes the earth spins, (it completes one full revolution in 24 hrs) - this is totally irrelevant. the only 'rotational' factor that applies is how the moon orbits the earth. It does so in an elliptical shape which I explained in my last post. which means you'll get a maximum force (at the perigee) and a minimum force (at the appogee).

Honestly, pullyour teacher up on this. Hold back after class and ask them to explain, they should be thrilled by a student wanting to learn! - I can't stand when they try to pull the wool over your eyes by acting like they know. Teachers/lecturers are not all knowing beings that have all the answers - and if they pretend to be, you have a bad teacher.

here's the solution: (best to be as informed as possible so you can have an informed discussion)

G = gravitational constant = 6.67 x 10^-11 m^3. kg^-1. s^-2
M_moon = Mass of the moon = 7.35 x 10^22 kg
M_earth = Mass of the earth = 5.97 x 10^27 kg
r_Perigee = minimum distance from earth to moon = 363295000 m
r_Apogee = maximum distance from earth to moon = 405503000 m

F_Perigee = (6.67 x 10^-11). (((7.35 x 10^22 )*(5.97 x 10^27 ))/(363295000)^2) = 2.21884 x 10^23 newtons
F_Apogee = (6.67 x 10^-11). (((7.35 x 10^22 )*(5.97 x 10^27 ))/(405503000)^2) = 1.78097 x 10^23 newtons

Trust me; i'm an aeronautical engineer :cool: lol

PS, the reason the moon doesn't crash into us and kill us all us is due to it's orbit around the earth, and the spin of the earth (as your teacher may have mentioned). These factors give the moon angular momentum in order to keep from crashing into us.

BUT, the question was about gravitational force. (as explained above)

basically, Gravity wants to bring the moon closer, angular momentum wont let it get closer.

toastedpickles

sebastianlieken wrote: »

the moon orbits the earth, the earth (and moon) orbit the sun, the sun is spinning in the milkey way, and the milkey way is wizzing through the universe, etc etc... that doesn't matter one bit when it comes to gravitational forces. A gravitational force is an attraction between 2 objects as far as you're concerned - when you start factoring in every other factor you're delving into quantum mechanics which is lunacy if you havn't even been thought the basics yet.

Yes the earth spins, (it completes one full revolution in 24 hrs) - this is totally irrelevant. the only 'rotational' factor that applies is how the moon orbits the earth. It does so in an elliptical shape which I explained in my last post. which means you'll get a maximum force (at the perigee) and a minimum force (at the appogee).

Honestly, pullyour teacher up on this. Hold back after class and ask them to explain, they should be thrilled by a student wanting to learn! - I can't stand when they try to pull the wool over your eyes by acting like they know. Teachers/lecturers are not all knowing beings that have all the answers - and if they pretend to be, you have a bad teacher.

here's the solution: (best to be as informed as possible so you can have an informed discussion)

G = gravitational constant = 6.67 x 10^-11 m^3. kg^-1. s^-2
M_moon = Mass of the moon = 7.35 x 10^22 kg
M_earth = Mass of the moon = 5.97 x 10^27 kg
r_Perigee = minimum distance from earth to moon = 363295000 m
r_Apogee = maximum distance from earth to moon = 405503000 m

F_Perigee = (6.67 x 10^-11). (((7.35 x 10^22 )*(5.97 x 10^27 ))/(363295000)^2) = 2.21884 x 10^23 newtons
F_Apogee = (6.67 x 10^-11). (((7.35 x 10^22 )*(5.97 x 10^27 ))/(405503000)^2) = 1.78097 x 10^23 newtons

Trust me; i'm an aeronautical engineer :cool: lol

Haha well look at you :P

We have him in a few hours, so i shall report back :pac:

Everyone cheered seeing you say we have a bad teacher, we thought it was just us :P

sebastianlieken

toastedpickles wrote: »

Haha well look at you :P

We have him in a few hours, so i shall report back :pac:

Everyone cheered seeing you say we have a bad teacher, we thought it was just us :P

"everyone", have I an audience or something?!

also, read the bottom of my last post. maybe that's what he was on about?

He said "gravitational force" thoough.

in terms of the angular moment that keeps the moon "up" , quick explanation:

you throw a bowling ball down the lane and it goes in a straight line, now give the ball a kick half way down the lane, it doesn't immediatly go into the gutter as it has momentum, instead it changes it's direction slightly.

if the earth suddenly disappeared, the moon would just drift off in a straight line like that bowling ball (ignore all other planets). The earth's graviational attraction is that "kick" that is making the moon change it's direction; and it does so continuously so that the moon orbits around the earth. dunno if that makes sense actually...

toastedpickles

sebastianlieken wrote: »

"everyone", have I an audience or something?!

also, read the bottom of my last post. maybe that's what he was on about?

He said "gravitational force" thoough.

in terms of the angular moment that keeps the moon "up" , quick explanation:

you throw a bowling ball down the lane and it goes in a straight line, now give the ball a kick half way down the lane, it doesn't immediatly go into the gutter as it has momentum, instead it changes it's direction slightly.

if the earth suddenly disappeared, the moon would just drift off in a straight line like that bowling ball (ignore all other planets). The earth's graviational attraction is that "kick" that is making the moon change it's direction; and it does so continuously so that the moon orbits around the earth. dunno if that makes sense actually...

So...we called him up on it, now he's changed his tune and told us to skip it :mad:
People!!

sebastianlieken

lol, what a w4nker! :P

toastedpickles

sebastianlieken wrote: »

lol, what a w4nker! :P

You have noooo idea :pac:

sebastianlieken

is this uni or secondary school by the way? and if uni, which one?

toastedpickles

sebastianlieken wrote: »

is this uni or secondary school by the way? and if uni, which one?

uni

DIT architectural technology

riffmongous

toastedpickles wrote: »

We still can't get it, the joys or structural mechanics eh? apparently his "logic" is that the moon and earth rotate together instead of the moon around the earth, We've given up

Although i learned more from you guys in 10 minutes than i have from this guy all semester

I think what he means here is that they orbit around their common centre of mass. Think of when you say the moon orbits around the earth.. it doesnt orbit around the centre of mass of the earth, it orbits around the common centre of mass of the system which happens to lie within the earth. Why is this important to know? Well its one of the techniques they first used to discover planets in other solar systems, if you have a large planet orbitining very close to a star, the common centre of mass can be slightly outside the star, meaning the star will have a small orbit that can be observed

Carawaystick

riffmongous wrote: »

I think what he means here is that they orbit around their common centre of mass. Think of when you say the moon orbits around the earth.. it doesnt orbit around the centre of mass of the earth, it orbits around the common centre of mass of the system which happens to lie within the earth. Why is this important to know? Well its one of the techniques they first used to discover planets in other solar systems, if you have a large planet orbitining very close to a star, the common centre of mass can be slightly outside the star, meaning the star will have a small orbit that can be observed

And that point is the Barycentre.

For Pluto/Charon minor planets, their barycentreis outside the surface of either dwarf planet.