(A) The kinetic energy $(KE)$ of a satellite in a circular orbit is given by the formula: $KE = \frac{GM_e m}{2r}$,where $r = R_E + h$.Given: $M_e = 6

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(A) The kinetic energy $(KE)$ of a satellite in a circular orbit is given by the formula: $KE = \frac{GM_e m}{2r}$,where $r = R_E + h$.Given: $M_e = 6 \times 10^{24} \ kg$,$m = 1000 \ kg$,$R_E = 6.4 \times 10^6 \ m$,$h = 270 \ km = 0.27 \times 10^6 \ m$,$G = 6.67 \times 10^{-11} \ Nm^2 \ kg^{-2}$.Calculating $r$: $r = 6.4 \times 10^6 + 0.27 \times 10^6 = 6.67 \times 10^6 \ m$.Substituting the values into the formula:$KE = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 6.67 \times 10^6}$$KE = \frac{6.67 \times 6 \times 10^{16}}{2 \times 6.67 \times 10^6} = 3 \times 10^{10} \ J$.Thus,the kinetic energy is $3 \times 10^{10} \ J$.

Class 11 Physics Gravitation Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits

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$A$ 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 $= 6.67 \times 10^{-11} \ Nm^2 \ kg^{-2}$)

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Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits

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Two satellites $A$ and $B$ are revolving around the earth in orbits of heights $1.25 R_E$ and $19.25 R_E$ from the surface of the earth respectively,where $R_E$ is the radius of the earth. The ratio of the orbital speeds of the satellites $A$ and $B$ is (in $: 1$)

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The largest and the shortest distance of the earth from the sun are ${r_1}$ and ${r_2}$ respectively. What is its distance from the sun when it is at a position perpendicular to the major axis of the orbit drawn from the sun?

DifficultAIPMT 1988

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Consider two satellites $S_{1}$ and $S_{2}$ with periods of revolution $1\, hr$ and $8\, hr$ respectively,revolving around a planet in circular orbits. The ratio of the angular velocity of satellite $S_{1}$ to the angular velocity of satellite $S_{2}$ is:

MediumJEE MAIN 2021

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Consider a binary star system consisting of two stars of masses $M_1$ and $M_2$ separated by a distance of $30 \text{ AU}$ with a period of revolution equal to $30 \text{ years}$. If one of the two stars is $5$ times farther from the centre of mass than the other, find the masses of the two stars in terms of solar masses.

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The distances of two planets $A$ and $B$ from the sun are $r_A$ and $r_B$ respectively. Given that $r_B = 100 r_A$. If the orbital speed of planet $A$ is $v$,then the orbital speed of planet $B$ is:

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Two satellites $A$ and $B$ are revolving around the earth in orbits of heights $1.25 R_E$ and $19.25 R_E$ from the surface of the earth respectively,where $R_E$ is the radius of the earth. The ratio of the orbital speeds of the satellites $A$ and $B$ is (in $: 1$)

The largest and the shortest distance of the earth from the sun are ${r_1}$ and ${r_2}$ respectively. What is its distance from the sun when it is at a position perpendicular to the major axis of the orbit drawn from the sun?

Consider two satellites $S_{1}$ and $S_{2}$ with periods of revolution $1\, hr$ and $8\, hr$ respectively,revolving around a planet in circular orbits. The ratio of the angular velocity of satellite $S_{1}$ to the angular velocity of satellite $S_{2}$ is:

The distances of two planets $A$ and $B$ from the sun are $r_A$ and $r_B$ respectively. Given that $r_B = 100 r_A$. If the orbital speed of planet $A$ is $v$,then the orbital speed of planet $B$ is:

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