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If a planet existed with orbital radius 4 times that of earth, how many years would it take to go around the sun once?

Posted by Meenu Biji 9 years ago

CBSE > Class 11 > Physics

1 answers

Ankit Kumar 9 years ago

as we know the expresssion for orbital velocity as,

$v_{0} = \sqrt{\frac{G M}{R + h}} w h e r e v_{0} = o r b i t a l v e l o c i t y R = r a d i u s o f s u n M = m a s s o f s u n R + h = o r b i t a l r a d i u s G = g r a v i t a t i o n a l c o n s \tan t$

R+h= $R_{0}$ (Let)

now the above equation can be writtenn as in terms of circumference and time i.e., $v_{0}$ = circumference made by earth / time taken by it.

so, $\frac{2 π (R + h)}{t} = \sqrt{\frac{G M}{R + h}} a s w e a l r e a d y a s l e t (R + h) = R_{0} s o \frac{2 π R_{0}}{t} = \sqrt{\frac{G M}{R_{0}}} t = \frac{2 π R_{0}}{\sqrt{\frac{G M}{R_{0}}}} n o w w h e n o r b i t a l r a d i u s i s 4 t i m e s o f o r b i t a l r a d i u s m a d e b y e a r t h i . e ., {R^{'}}_{0} = 4 R_{0} l e t t^{'} b e t h e t i m e m a d e b y u n k n o w n p l a n e t s o t^{'} = \frac{2 {πR}^{'}_{0}}{\sqrt{\frac{G M}{{R^{'}}_{0}}}} i m p l i e s t h a t t^{'} = \frac{2 π 4 R_{0}}{\sqrt{\frac{G M}{4 R_{0}}}} n o w \frac{t}{t^{'}} = \frac{\frac{2 π R_{0}}{\sqrt{\frac{G M}{R_{0}}}}}{\frac{2 π 4 R_{0}}{\sqrt{\frac{G M}{4 R_{0}}}}} o n s o l v i n g t h e a b o v e e x p r e s s i o n w e g e t t^{'} = 8 t h e n c e t h e r e q u i r e d t i m e i s t^{'} = 8 X 1 y e a r s = 8 y e a r s$