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# Gravitation questions (Help plss!!) watch

1. Yo all hope you all can help me on this q

1)The average radius for pluto`s orbit is 39.4AU and its period is 9.04 x 10^4 days.However,the irregularity in pluto`s orbit has been observed.There has been speculation that ,In order to explain such irregularity,a tenth plane X must be present.
If X were to exist,and if its period is twice to that of pluto,calculate at what average distance would X would orbit around the sun.

Note AU = 1.5 x 10^11 m
2. 2nd question
given that the diameter of pluto is 6 x 10^6 m
If an observer were to be standing on a point O at planet X
what would be tha angle of the observer`s sight range on the surface of pluto
3. 1. 9.38x10^12m = 62.5 AU

Is this right?
4. (Original post by rts)
1. 9.38x10^12m = 62.5 AU

Is this right?
thats aboslutely right man
how did you do it
and can you help me find part 2?
5. (Original post by MalaysianDude)
thats aboslutely right man
how did you do it
and can you help me find part 2?
Don't you need to know the distance between Pluto and planet X? Then you can use α/206265" = d/r to work out α in arcsec (and then convert to radians/degrees). d is the linear diameter and r is the distance between Pluto and X.
6. (Original post by Nylex)
Don't you need to know the distance between Pluto and planet X? Then you can use α/206265" = d/r to work out α in arcsec (and then convert to radians/degrees). d is the linear diameter and r is the distance between Pluto and X.
i dont think you need it
and btw,i got this q from an OCR past year paper and found it pretty hard to solve...

and if rts could get the distance,i think its possible to solve it wouldnt it?

and can someone get the answer for q2??
7. 1. t = period of planet X
T = period of Pluto
R = mean radius of orbit of Pluto
r = mean radius of orbit of planet X

By Kepler's Law:

T^2/R^3 = t^2/r^3 = (2T)^2/r^3 = 4T^2/r^3

therefore r^3 = 4R^3

r = 62.5AU
8. (Original post by rts)
1. t = period of planet X
T = period of Pluto
R = mean radius of orbit of Pluto
r = mean radius of orbit of planet X

By Kepler's Law:

T^2/R^3 = t^2/r^3 = (2T)^2/r^3 = 4T^2/r^3

therefore r^3 = 4R^3

r = 62.5AU
omg..keplar`s law?
never heard of it man...
is it derived from newton`s law of gravitation and the centripetal force?

and also...can you pls do number 2?
9. Yeah, Kepler's 3rd law is derived from Newton's law of gravitation (taking the gravitational force as being a centripetal force).
10. (Original post by MalaysianDude)
omg..keplar`s law?
never heard of it man...
is it derived from newton`s law of gravitation and the centripetal force?

and also...can you pls do number 2?
mv^2/R = GMm/R^2

v = root(GM/R)

T = 2*pi*R/v = 2*pi*R/root(GM/R) = 2*pi*R*root(R)/root(GM)

T^2 = 4*pi*pi*R^3/GM

T^2 = kR^3 (Mass of sun is constant)

T^2/R^3= k. Kepler's Third Law.

I have no idea how to do number two. Never done astronomy before.
11. Dude,this isnt astronomy
more like a normal gravitation question they just putplanets to scare the **** out of you
so pls do try how to do it..
i would guess that if you take the value of the distance of planet X from the sun and distance of pluto from the sun.
Deduct both of these then use tanX to derive them
but i still couldnt get it
12. is this maths / physics?
13. (Original post by ShOcKzZ)
is this maths / physics?
physics
14. (Original post by MalaysianDude)
Dude,this isnt astronomy
more like a normal gravitation question they just putplanets to scare the **** out of you
so pls do try how to do it..
i would guess that if you take the value of the distance of planet X from the sun and distance of pluto from the sun.
Deduct both of these then use tanX to derive them
but i still couldnt get it
Is the angle = 1*10^-6 degrees ?
15. (Original post by Fermat)
Is the angle = 1*10^-6 degrees ?
erm
thats not the right answer cause its in degrees
16. (Original post by MalaysianDude)
erm
thats not the right answer cause its in degrees
Rx = radius of planet X
L = distance between them
D = diameter of pluto

Rx=62.5 Au
Rp=39.4 Au
L = Rx - Rp
L = 62.5 - 39.4
L = 23.1 Au
L = 3.465 x 10^12 m
D = 6 x 10^6 m

tan ø = (D/2)/L
tan ø = (3 x 10^6)/(3.465 x 10^12)
tan ø = 8.66 x 10^-7
ø = 8.66 x 10^-7 rads
2ø = 1.732 x 10^-6 rads
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