YumYum247
  • YumYum247
Kelp me Please!!!! :")
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
can someone please check the Net force between the person (65kg) and the earth with a mass of 5.98X10^24Kg and radius of 6.38X10^6m. And how do i find the net force between them the second way...i'm not getting the right answer!!!
YumYum247
  • YumYum247
@PlasmaFuzer Please help me!!!! :")
YumYum247
  • YumYum247
is it.....|dw:1442282566185:dw|

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anonymous
  • anonymous
Hello whats going on :D
YumYum247
  • YumYum247
nothing much, just another day stuck on physics as usual :D
YumYum247
  • YumYum247
i think i got it.... but i don't quite know how!!!! :"D
YumYum247
  • YumYum247
i was suppose to get the net force between the earth and the person....i got it the first way by applying the conventional formula.... F = G(m1)(m2)/d^2 and got 635N
YumYum247
  • YumYum247
sorry i forgot to tell you...it's question "b"
YumYum247
  • YumYum247
and the other way i got the net force was by multiplying the mass off the person with the gravitational force of wearth...9.80N and i got 637N
YumYum247
  • YumYum247
but what do you call that in terms of an actual formula...?!?!?!?
anonymous
  • anonymous
Ok so the way I would do it is as follows: Since the force is the same (the force of gravity I can cheat and use the ratio of the two.... On Earth F=weight=mg=(65kg)g On Jupiter F'=jupiter weight=mj = (65kg)j Now divide the equations which give you F/F' = g/j
anonymous
  • anonymous
So clearly j=g*(F'\F)
YumYum247
  • YumYum247
My dear what is F' and m/j, i'm not familiar to those variables.....:"(
YumYum247
  • YumYum247
i'm sorry if that makes things difficult....but i don't know what they mean...!!! :"(
anonymous
  • anonymous
Ok sorry I am hoping around to much
YumYum247
  • YumYum247
did i do part b right..?!?!?!? please check ")
anonymous
  • anonymous
Oh that I defined F to be the force of gravity on earth (the weight on earth, F' to be the weight on jupiter and g is the earth grav accel and j is the jupiter grav accel (just made that letter up)
YumYum247
  • YumYum247
OMG.....i'm sorry, i'm really dumb....please don't mind that :")
anonymous
  • anonymous
Ahhh no your'e part be isnt TOTALLY correct. If you put jupter's mass in a ball with the radius the size of the galaxy it's gravitational acceleration on the surface wouldn't be close to what it is on earth's surface.... what matters is that its Mass/Radius^2 which is honestly why I chose the approach I did...
anonymous
  • anonymous
Oh wait I read the question a bit backwards but my reasoning is still sound... The smaller planet here (earth) definitely DOES NOT have more mass than jupiter... google it ~10^24kg for earth and ~10^27 for jupiter
YumYum247
  • YumYum247
so your saying it's partly correct?!?!! ?O_oo_O
anonymous
  • anonymous
No now that I reread the question it is totally incorrect sorry :/
YumYum247
  • YumYum247
Aaawwnnn!!!! :"(
anonymous
  • anonymous
I just glanced at it and thought it was saying the weight on jupiters "surface" was greater than that on earth
anonymous
  • anonymous
but anways i digress
YumYum247
  • YumYum247
so what have i done wrong dear?!??! Are you sure you read the right part....question A is on the left side of the paper and B is on the right side....i'm sure B is correct because i've double checked it....
anonymous
  • anonymous
Ok so we had j=g*(F'/F)
YumYum247
  • YumYum247
tow different ways and i', getting the right answer@@@
anonymous
  • anonymous
Ok you will see why I do it this way in a second... it saves having to calculate a lot of numbers
YumYum247
  • YumYum247
please proceed your highness!!! :")
anonymous
  • anonymous
\[j=g \frac{F_{G_{Jupiter}}}{F_{G_{Earth}}}=g\frac{\frac{GmM_{Jup}}{R_{Jup}^2}}{\frac{GmM_{Earth}}{R_{Earth}^2}} = g \frac{\frac{M_{Jup}}{R_{Jup}^2}}{\frac{M_{Earth}}{R_{Earth}^2}}= g \frac{M_{Jup}R_{Earth}^2}{{R_{Jup}^2M_{Earth}}} \]
anonymous
  • anonymous
So now we have removed all those pesky extra factors... and have expressed it in terms of what we know
anonymous
  • anonymous
Awwww darn I just realized they dont give you the earth mass so you have to go through all the lame calculations..... (thinking) hold on
YumYum247
  • YumYum247
but it looks scary tho :(
anonymous
  • anonymous
Actually I can fix it hold on:
YumYum247
  • YumYum247
please don't fix it, let it be..i have a habit following those lame and long calculations....
anonymous
  • anonymous
Oh wait... were you allowed to just look up the values of the radius and mass of the earth?
anonymous
  • anonymous
because if so... all you need to do is take that equation plug in your knowns and evaluate it.... then of course you have found the acceleration on the surface of jupiter (I called it j)... then multiply by the mass to find the weight on the "surface" of jupiter
YumYum247
  • YumYum247
they were given to me in the question.
anonymous
  • anonymous
Nope I only see Jupiter's mass and radius as givens
anonymous
  • anonymous
But u were definitely using earth's mass and radius in one of the formulas
YumYum247
  • YumYum247
that's what i did honey.... the second way to get the net force between the person and the earth is by simply multiplying the gravitational force of gravity of earth(9.8m/sec^2) times the mass of the person...65kg...and i get the same answer as in the initial way i got the answer!!!
YumYum247
  • YumYum247
please ignore the last line on the paper...my second way there is wrong...........i made a mistake there....
anonymous
  • anonymous
Ummm yes part a) has you find the earth weight which is just mg but you did an F_grav calculation with earth's mass and radius which weren't givens... look if your'e teacher is ok with that I am I just wasn't sure because sometimes you have to stick to the knowns in the problem.
anonymous
  • anonymous
Its alright
YumYum247
  • YumYum247
please ignore the last line on the paper...my second way there is wrong...........i made a mistake there....
anonymous
  • anonymous
So your answer looks correct using the numbers you used int he problem 24.585 which you should express as 24.6 since the largest number of sig figs used in the givens is 3
anonymous
  • anonymous
I think thats how it works... it might be the minimum number of sig figs, but I think I have it right
anonymous
  • anonymous
sorry 24.585 N and 24.6 N shame on me for forgetting units
YumYum247
  • YumYum247
btw were you referring to art B, because the mass of the earth and the radius were given in the previouse page o i used those to solve for the net force....for part A i on'y used what was given to me...
anonymous
  • anonymous
OOOOO ok then yes then my way totally works... and yes I calculated the answer to part b) using my method... Note how it looks nasty, but I got rid of the extraneous factors before evaluating it
YumYum247
  • YumYum247
but the question is still there.... do you see the 635N on the paper...????? how do i get the same net force using another way.....
anonymous
  • anonymous
Given the level of accuracy required... it probably doesn't make any difference, but in the real world a scientific calculation should be first expressed as simply as possible with as few variables as possible and then evaluated. Since each term bring with it uncertainties which stack up (though some known values can have absurdly high accuracy like charge/mass ratio of electron)
YumYum247
  • YumYum247
^^.agree
anonymous
  • anonymous
oh... ummm I thought it was calculate earth's weight and then calculate jupiter's weight two different ways
YumYum247
  • YumYum247
no just the earth's.....
anonymous
  • anonymous
So the first way is definitely using F_gravity in its full form... then just use mg as the second one
YumYum247
  • YumYum247
there is another way to find the net force between two objects.....
anonymous
  • anonymous
heck all g represents is this: \[ g= \frac{GM_{Earth}}{R_{Earth}^2} \]
YumYum247
  • YumYum247
i think it's 9.80N X mass of the person (65kg)
YumYum247
  • YumYum247
yah...that's it....i get 637N which is pretty close to 635N....
anonymous
  • anonymous
So if you evaluate it long hand by using all those terms... and then evaluate it using just g technically they are the same but I dont know.... No that is incorrect the units of g are acceleration not force ( m/s/s)
YumYum247
  • YumYum247
i know but i get the right number tho :"(
anonymous
  • anonymous
Correct... but your teacher would also point that out. Units are important
YumYum247
  • YumYum247
F = mXa a = F/m i know :"(
anonymous
  • anonymous
I know its ok... just a simple error, but I can't let it go
anonymous
  • anonymous
Anyways I have to go and get some food so I will see you next time
YumYum247
  • YumYum247
so is it right or wrong?!?!?!
anonymous
  • anonymous
You calculated it two different ways and came to pretty much the same answer what do you think? XD
anonymous
  • anonymous
Im not trying to be rude, but since your answers agreed that would suggest they are probably correct.
YumYum247
  • YumYum247
ok Thank you and love you infinite....GOD bless you :")
YumYum247
  • YumYum247
Please take care of yourself we need more people like you here on this planet :") Live long.
YumYum247
  • YumYum247
btw it was on the spot guess kinda thing.....no joke.!! :D
anonymous
  • anonymous
:D your welcome... remember as you progress in your studies you will find it harder and harder for people to be able to help you and then maybe, eventually you will be out there on your own solving problems that don't have known answer. This is a really good time to start practicing your judgement and reasoning skills on problems that are easy, and when you do make an error to learn from it by finding out where you made the error and always being cautious of it down the road.
YumYum247
  • YumYum247
Again tHank you shoooooooooooooooooooo much!!! :")
anonymous
  • anonymous
:D cya later
YumYum247
  • YumYum247
Abolutely agree with you....:") Anyways Cya amigo :)

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