Gravitation questions (Help plss!!)

OP

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

Reply 2

rts

1. 9.38x10^12m = 62.5 AU

Is this right?

Reply 3

OP

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?

Reply 4

Nylex

10

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.

Reply 5

OP

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??

Reply 6

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

Reply 7

OP

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?

Reply 8

Nylex

10

Yeah, Kepler's 3rd law is derived from Newton's law of gravitation (taking the gravitational force as being a centripetal force).

Reply 9

rts

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.

Reply 10

OP

Dude,this isnt astronomy
more like a normal gravitation question they just putplanets to scare the shit 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

Reply 11

ShOcKzZ

is this maths / physics?

Reply 12

OP

ShOcKzZ

is this maths / physics?

physics

Reply 13

Fermat

8

MalaysianDude

Dude,this isnt astronomy
more like a normal gravitation question they just putplanets to scare the shit 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 ?

Reply 14

OP