TS EAMCET · Physics · Gravitation
If the orbital speed of a body revolving in a circular path near the surface of the earth is \(8 \mathrm{kms}^{-1}\), then the orbital speed of a body revolving around the earth in a circular orbit at height of \(19,200 \mathrm{~km}\) from the surface of earth is (Radius of the earth \(=6400 \mathrm{~km}\) )
- A \(4 \mathrm{kms}^{-1}\)
- B \(6 \mathrm{kms}^{-1}\)
- C \(7.5 \mathrm{kms}^{-1}\)
- D \(9 \mathrm{kms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(4 \mathrm{kms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(v_2 = v_1 \sqrt{\frac{R_E}{R_E + h}}\) \(v_2 = 8 \mathrm{kms}^{-1} \sqrt{\frac{6400 \mathrm{~km}}{6400 \mathrm{~km} + 19200 \mathrm{~km}}}\) \(v_2 = 8 \mathrm{kms}^{-1} \sqrt{\frac{6400 \mathrm{~km}}{25600 \mathrm{~km}}}\) \(v_2 = 8 \mathrm{kms}^{-1} \sqrt{\frac{1}{4}}\)…
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