A satellite launched with velocity V\(_E\) just escapes the earth's gravitational attraction. Given that the...

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 = \(\sqrt{2gR}\)

= \(\sqrt{20R}\)