A satellite of mass M/2 is revolving around earth in a circular orbit at a height of R/3 from earth surface. The angular momentum of the satellite is M√(GMR/x). The value of x is _________, where M and R are the mass and radius of earth, respectively. ( G is the gravitational constant) Numerical answer type. Work it out, then open the answer.

Answer: 3. Orbit radius r = R + R/3 = 4R/3, satellite mass M/2. L = m√(GMr) = M/2√(GM·4R/3) L = M√(1/4·4GMR/3) = M√(GMR/3) So x = 3.

A satellite of mass M/2 is revolving around earth in a circular orbit… | JEE Main 2025 (23 Jan, Shift 2)

Q16JEE Main 2025 · 23 JAN · SHIFT 2Medium

Satellites - Orbital Velocity & Period

A satellite of mass $\frac{M}{2}$ is revolving around earth in a circular orbit at a height of $\frac{R}{3}$ from earth surface. The angular momentum of the satellite is $M\sqrt{\frac{GMR}{x}}$ . The value of $x$ is _________, where M and R are the mass and radius of earth, respectively. ( G is the gravitational constant)

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Orbit radius $r = R + \frac{R}{3} = \frac{4R}{3}$ , satellite mass $\frac{M}{2}$ .
$L = m\sqrt{GMr} = \frac{M}{2}\sqrt{GM\cdot\frac{4R}{3}}$
$L = M\sqrt{\frac{1}{4}\cdot\frac{4GMR}{3}} = M\sqrt{\frac{GMR}{3}}$
So $x = 3$ .

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