Question

an undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. this moon takes 1960*10^3 s(about 23 days) on average to complete one nearly circular revolution around the unnamed planet. if the distance from the center of the moon to the surface of the planet is 215.0*10^6m and the planet has a radius of 3.40*10^6m, calculate the moon's radial acceleration.

i've used 2(pi)/t...is that right??

Answer 1

2pi/T will give you the angular velocity of the moon

Answer 2

so do i use a = v²/R?

Answer 3

a+v^2/R?

Answer 4

*=...not +

Answer 5

well if you've already worked out the angular velocity, call it omega, why not use the formula r x (omega squared) for the radial acceleration

Answer 6

v squared over r is correct, but you may as well work with the omega formula

Answer 7

\[a _{r}=v^2/r=r \omega^2\]
either way is fine

Answer 8

do i have to find the total distance first? i did that and got= 2.184*10^8
then i did 2(pi)/t t=1960*10^3 and got =3.21/s
i then entered that into a=Rw^2 and got 2250.4*10^6 as the final answer....am i doing it correct?

Answer 9

first, yes you are right, the radius we need is the sum of the planet radius plus the surface to moon distance

you're calculation of omega is not quite correct - you are missing an exponent

Answer 10

\[\frac{ 2\pi }{ 1960e10^3 }=?\]

Answer 11

.0644...right?

Answer 12

erm, no
2pi/1960000

Answer 13

oh so it'll be 3.206e^-6?

Answer 14

correct
now you just need to calculate r times omega squared for the radial acceleration

Answer 15

so i should get 55.64*10^-5?

Answer 16

what do you have for r ?

Answer 17

the total distance...

Answer 18

yes, the distance from centre of planet to centre of moon

Answer 19

yes it was 218.4*10^6....thats supposed to be r right?

Answer 20

that is right, yes
and what do you have for omega squared ?

Answer 21

3.06e^-6

Answer 22

that is omega

Answer 23

yes..so squared it would be 10.28e^-6?

Answer 24

i don't think so

Answer 25

what is 10^-6 squared ?

Answer 26

you can't just square the first few numbers and forget about the power of ten, that has to be squared as well !

Answer 27

oh ok, so it would be 1.97e^-5

Answer 28

what would be ?

Answer 29

omega

Answer 30

omega we already calculated to be 3.2e10-6

Answer 31

now we need omega squared

Answer 32

r= 218.4*10^6 and omega squared =1.97e^-5

Answer 33

???????????????????????

Answer 34

omega square = omega multiplied by omega
if omega is 3.2e10-6, just work out omega squared for me

Answer 35

i get 1.024e^13

Answer 36

you need practise doing simple calculations with numbers in scientific notation
if omega = 3.2e10-6 then omega squared = 1.024e10-11

Answer 37

you have to take great care to be accurate doing these numerical calculations

Answer 38

where are you getting e10-11 from?

Answer 39

i'm just entering it into my calculator

Answer 40

the power of 10 in omega is minus 6
if you square it, that will give you a power of minus 12, but 3.2 squared is about 10, so instead of writing 10 times 10 to power -12, you write 1 times 10 to power -11

Answer 41

oooh ok, i see now

Answer 42

it's very important to be able to do that kind of calculation correctly, otherwise you might understand the physics but not be able to get the correct numbers out

Answer 43

so now you have r and you have omega squared - can you calculate the radial accereration for me ?

Answer 44

i have 82273146.83

Answer 45

oh my god, where did that come from

Answer 46

r = 218.4 x 10^6 and omega squared = 1.02 x 10^-11, just multiply those two numbers together

Answer 47

i hit the wrong button, sorry..the correct answer is .00222768

Answer 48

that's more like it : )
so the radial acceleration is just over 2mm per second squared.

Answer 49

yes..thanks so much, seems i'm making careless mistakes in my calculations

Answer 50

it's easily done - a good night's sleep can help

Answer 51

yes that's true i haven't slept in couple of days now, thanks again

Answer 52

welcome