Question

find the total energy and binding energy of an artificial satellite of mads 800 kg orbiting at a height of 1800 km above the surface of the earth.
When G = 6.67×10(-11) ai units. Radius of Earth = 6400km
mass of Earth M = 6×10²⁴ kgl​

Answer 1

Answer:

total energy ....1.952×10 power 10 J

binding energy also same as total energy...

Answer 2

$\huge\boxed{ \underline{ \underline{ \bf{Answer}}}}$

$==>$ E = $-\:1.952\:*\: 10^{10} \:J$

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${ \underline{ \underline { \bf{Given\::-}}}}$

M = 800 kg

h = 1800 km = $18\:*\:10^{5}$m

G = $6.67$×$10^{-11}$ S.I unit.

R = 6400 Km = 64 × $10^{5}$ m

M = $6$×$10^{24}$ kg.

We have to find out ,

E = ?   B.E = ?

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${ \underline{ \underline { \bf{Explanation\::-}}}}$

∵ B.E =  $\frac{GMm}{2\:(R+h)}$............( 1 )

         =  $\frac{6.67\:*\:10^{-11}*6\:*\:10^{24}\:*\:800}{2[64\:*\:10^{5}+18\:*\:10^{5}]}$

         = $\frac{6.67\:*\:6\:*\:8\:*\:*\:10^{15} }{2\:*\:82\:*\:10^{5} }$

         =  $\frac{6.67\:*\:48}{164} \:*\:10^{10}$

∴B.E =  $1.952\:*\:10^{10} J$

Also,        

E       = $-\:\frac{GMm}{2\:(R+h)}$............(2)

now, compare Equation (1) & (2) we found,

E      =  - B.E

E     = $-\:1.952\:*\: 10^{10} \:J$

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