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 ______ $\times 10^{10} \mathrm{~J}$. (Mass of earth $=6 \times 10^{24} \mathrm{~kg}$, Radius of earth $= 6.4 \times 10^{6} \mathrm{~m}$, Gravitational constant $= 6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$)

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Gravitation

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 1 0^{10} J

.
(Mass of earth

= 6 \times 1 0^{24} kg

, Radius of earth

= 6.4 \times 1 0^{6} m

, Gravitational constant

= 6.67 \times 1 0^{- 11} Nm^{2} kg^{- 2}

)

Answer: 3

Solution

For a circular orbit the gravitational pull provides the centripetal force,

\frac{G M _{e} m}{r ^{2}} = \frac{m v ^{2}}{r}

, so the kinetic energy is

K E = \frac{1}{2} m v^{2} = \frac{G M _{e} m}{2 r}

. The orbital radius is

r = R_{E} + h = 6.4 \times 1 0^{6} + 2.7 \times 1 0^{5} = 6.67 \times 1 0^{6}

m. Then

K E = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 6 \times 1 0 ^{24} ) ( 1000 )}{2 ( 6.67 \times 1 0 ^{6} )} = \frac{4.002 \times 1 0 ^{17}}{1.334 \times 1 0 ^{7}} = 3 \times 1 0^{10}

J. The kinetic energy of the satellite is

3 \times 1 0^{10}

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Gravitation

Solution

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