A satellite launched with velocity V\(_E\) just escapes the earth's gravitational attraction. Given that the radius of the earth is R, show that V\(_E\) = \(\sqrt{20R}\) [g = 10ms\(^{-2}\)
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Correct Answer: Option
Explanation:
To show that V\(_E\) = \(\sqrt{20R}\)
KE = Gravitational Potential Energy
\(\frac{1}{2}\)MV\(_E^2\) = \(\frac{GmM}{R}\)
V\(_E^2\) = \(\frac{2GM}{R}\)
but GM = gR\(^2\)
V\(_E^2\) = \(\frac{2gR^2}{R}\) = 2gR
V\(_E^2\) = \(\sqrt{2 \times 10 \times R}\)
= \(\sqrt{20R}\)
OR
P.E = K.E
mgR = \(\frac{1}{2}\)mv\(^2\)
v\(^2\) = 2gR
v = \(\sqrt{2gR}\)
= \(\sqrt{20R}\)
To show that V\(_E\) = \(\sqrt{20R}\)
KE = Gravitational Potential Energy
\(\frac{1}{2}\)MV\(_E^2\) = \(\frac{GmM}{R}\)
V\(_E^2\) = \(\frac{2GM}{R}\)
but GM = gR\(^2\)
V\(_E^2\) = \(\frac{2gR^2}{R}\) = 2gR
V\(_E^2\) = \(\sqrt{2 \times 10 \times R}\)
= \(\sqrt{20R}\)
OR
P.E = K.E
mgR = \(\frac{1}{2}\)mv\(^2\)
v\(^2\) = 2gR
v = \(\sqrt{2gR}\)
= \(\sqrt{20R}\)