JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2026
- Hard
- MCQ
Initially a satellite of 100 kg is in a circular orbit of radius \(1.5 R _{ E }\). This satellite can be moved to a circular orbit of radius \(3 R _{ E }\) by supplying \(\alpha \times 10^6 J\) of energy. The value of \(\alpha\) is _________ .
(Take Radius of Earth \(R_E=6 \times 10^6 m\) and \(g =10 m / s ^2\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 1000
Step-by-step Solution
Detailed explanation
Energy of a satellite in a circular orbit is given as\(E =\frac{- GM _{ E } m }{2 r } ; r =\) radius of circular orbit
Required energy to be supplied \(=\Delta E=E_f-E_i\)
\(\Delta E =\left(\frac{- GM _{ E } m }{2\left(3 R _{ E }\right)}\right)-\left(\frac{- GM _{ E } m }{2\left(1.5 R _{ E }\right)}\right)\)
\(=\frac{ GM _{ E } m }{6 R _{ E }}\)
Now, \(g =\frac{ GM _{ E }}{ R _{ E }^2} \Rightarrow \frac{ GM _{ E }}{ R _{ E }}= gR _{ E }\)
\(\therefore \quad \Delta E =\frac{1}{6} gmR _{ E }\)
\(=\frac{1}{6} \times 10 \times 100 \times 6 \times 10^6\)
\(=1000 \times 10^6\)
\(\alpha=1000\)
Required energy to be supplied \(=\Delta E=E_f-E_i\)
\(\Delta E =\left(\frac{- GM _{ E } m }{2\left(3 R _{ E }\right)}\right)-\left(\frac{- GM _{ E } m }{2\left(1.5 R _{ E }\right)}\right)\)
\(=\frac{ GM _{ E } m }{6 R _{ E }}\)
Now, \(g =\frac{ GM _{ E }}{ R _{ E }^2} \Rightarrow \frac{ GM _{ E }}{ R _{ E }}= gR _{ E }\)
\(\therefore \quad \Delta E =\frac{1}{6} gmR _{ E }\)
\(=\frac{1}{6} \times 10 \times 100 \times 6 \times 10^6\)
\(=1000 \times 10^6\)
\(\alpha=1000\)
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