A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ___ . (Mass of earth , Radius of earth , Gravitational constant ) [20

Question

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ___ $\times 10^{10} J$ . (Mass of earth $= 6 \times 10^{24} kg$ , Radius of earth $= 6.4 \times 10^{6} m$ , Gravitational constant $= 6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}$ ) [2025]

Smart Achievers · Accepted Answer

(3)

$K = - \frac{U}{2} = \frac{G M m}{2 r}$

$\Rightarrow K = \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 10^{3}}{2 \times 6670 \times 10^{3}} = 3 \times 10^{10} J$

Solution

Ans.
(3)

$K = - \frac{U}{2} = \frac{G M m}{2 r}$

$\Rightarrow K = \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 10^{3}}{2 \times 6670 \times 10^{3}} = 3 \times 10^{10} J$

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Solution

Ans. (3) K=–U2=GMm2r ⇒ K=6.67×10–11×6×1024×1032×6670×103=3×1010J

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Ans.
(3)

$K = - \frac{U}{2} = \frac{G M m}{2 r}$

$\Rightarrow K = \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 10^{3}}{2 \times 6670 \times 10^{3}} = 3 \times 10^{10} J$