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Solar System Data

by aidan
Tags: data, solar
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aidan
#1
Jul18-03, 04:41 AM
P: n/a
Evening all.
My teacher has given us the mass, distance, and gravitational force of every planet in the solar system. we have to work out every planets acceleration, speed, and orbital period.
I'm having some trouble with the acceleration. This sshould be a very small number right?
I'm using the equation f=ma. (This is for Venus)
f=ma
a=f/m

f=1.12*10^34
m=4.89*10^24

now when i do f/m i get 2,290,388,548. Thats a huge number and I'm sure that Venus cant be accelerating that much.
My answer for Mercury was 0.0246 m/s.

Please help!!!
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Janus
#2
Jul18-03, 09:28 AM
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Your value for f is way too high. How did you arrive at this number?
S.P.P
#3
Jul18-03, 12:35 PM
P: 39
I think the formula you want for the acceleration is (v^2)/R where v is velocity and R is the radius of the planets orbit. Hope that helps.

Janus
#4
Jul18-03, 02:32 PM
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Solar System Data

Originally posted by S.P.P
I think the formula you want for the acceleration is (v^2)/R where v is velocity and R is the radius of the planets orbit. Hope that helps.
There's nothing wrong with his method, just the starting numbers.

For instance, with your method, he would first have to get the planet's velocity.

given that centripetal force = Mv/r,

v= [squ] fr/M) =

(using the numbers given)

[squ](1.12 e34 * 1.082e11/4.89e24 = 1.57e 10 m/s

plug this into v/r gives

(1.57e10)/1.082e11 = 2286154928

Which is of the same order as the answer he already has.
aidan
#5
Jul19-03, 03:08 AM
P: n/a
i got f doing this:

f = G.m1.m2 / r^2

G = 6.67*10^-11
m1 = 1.99*10^30
m2 = 4.89*10^24
r = 1.08*10^11

which equals, = ...5.56*10^22

is this better??
Janus
#6
Jul19-03, 11:02 AM
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Originally posted by aidan
i got f doing this:

f = G.m1.m2 / r^2

G = 6.67*10^-11
m1 = 1.99*10^30
m2 = 4.89*10^24
r = 1.08*10^11

which equals, = ...5.56*10^22

is this better??
Yes.


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