FIND THE NEW VALUE OF THE, ACCELERATION DUE TO GRAVITY IF THE RADIUS OF THE EARTH IS REDUCED BY 10%,WHHERE DENSITY OF EARTH REMAINS CONSTANT?

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Solution

$N o w, \sin c e d e n s i t y r e m a i n s c o n s \tan t a n d t h e r a d i u s d e c r e a s e s b y 10 %, i . e ., b e c o m e s 0.9 t i m e s t h e o r i g i n a l r a d i u s, t h e r e f o r e, R_{2} = 0.9 R_{1} n o w, d e n s i t y i s c o n s \tan t, S o l e t t h e n e w m a s s b e M_{2} a n d M_{1} b e t h e i n i t i a l m a s s t h e r e f o r e, \frac{M_{1}}{\frac{4}{3} {R_{1}}^{3}} = \frac{M_{2}}{\frac{4}{3} {R_{2}}^{3}} \frac{M_{1}}{\frac{4}{3} {R_{1}}^{3}} = \frac{M_{2}}{\frac{4}{3} (0.9 R_{1})^{3}} \frac{M_{1}}{\frac{4}{3} {R_{1}}^{3}} = \frac{M_{2}}{\frac{4}{3} \times 0.729 \times {R_{1}}^{3}} M_{2} = 0.729 M_{1} - - - - - - - - - - - - (1) n o w, \sin c e, a c c e l e r a t i o n d u e t o g r a v i t y i s g i v e n a s u n d e r : g = G \frac{M_{E}}{{R_{E}}^{2}} t h e r e f o r e, n e w a c c e l e r a t i o n d u e t o g r a v i t y i s g i v e n a s u n d e r g^{'} = G \frac{{M_{E}}^{'}}{({R_{E}}^{'})^{2}} w h e r e {M_{E}}^{'} = 0.729 M_{E} a n d {R_{E}}^{'} = 0.9 R_{E} g^{'} = G \frac{M_{E}}{(R_{E})^{2}} \times \frac{0.729}{0.81} g^{'} = 0.9 g g^{'} = 0.9 \times 9.81 g^{'} = 8.829 m s^{- 2}$

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