Question

The planet Jupiter of mass 2x10^27 kg revolves round the sun of mass 2x10^30 kg in a circular orbit of radius 7.8x10^11 m ,Calculate the gravitational force and orbit speed of Jupiter

Answer 1

i've calculated the gravitational force
its coming F=4.38x10^23

Answer 2

but what about orbit speed of jupiter ??

Answer 3

oh its circular

Answer 4

root mew over radius

Answer 5

\[V \approx \sqrt{\frac{ \mu }{ r }}\]

Answer 6

Sorry mew is just the gravitational constant 6.67 x10^-11 multiplied by the mass of whatevers curving it

Answer 7

the answer given in my book is 1.3077x10^4 m/s

Answer 8

the formula is given wrong answer

Answer 9

i was wrong, sorry, orbits are elliptical in most cases, you take the formula from the first eqation, equate it with kinetic energy, as Jupiter is a huge classical body, and rearrange for V

Answer 10

well because Jupiter is non-relativistic

Answer 11

rr \[1/2mv^2 = Gmm/r^2 \]

Answer 12

x 2, /m, root

Answer 13

1/2 mv^2 = 4.38x10^23

Answer 14

1/2*(2x10^27)*v^2 = 4.38x10^23

Answer 15

(1x10^27)v^2=4.38x10^23

Answer 16

v^2=4.38x10^-4

Answer 17

v=0.02092 m/s

Answer 18

lol

Answer 19

the answer is wrong again .... the real one is 1.3077x10^4

Answer 20

the first formula i gave you is correct

Answer 21

\[V = \sqrt{\frac{ 6.67 \times 10^{-11} \times 2\times10^{30}) }{ 7.8 \times 10^{11}}\]

Answer 22

That hasnt converted the LaTex on my screen, do you understand?

Answer 23

i cannot see what have u've posted

Answer 24

use the first formula, but the big M is the dominant sphere of influence, namely The Sun.

Answer 25

mew = G times M

Answer 26

The gravitation force from the sun to jupiter is the centripetal force.\[4,38 10^{23}=\frac{ m _{jup} }{ R}u ^{2}\]

Answer 27

oh my god of course

Answer 28

derivation is right now, thanks