An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is x(GMe m)/(21Re). The value of x is (take Re = 6370 km): (a) 12 (b) 11 (c) 9 (d) 10

Answer: (b). h = 318.5 km = Re/20, so r = (21Re)/(20). In orbit: Ef = -(GMem)/(2r) = -(10GMem)/(21Re). On the surface (at rest): Ei = -(GMem)/(Re) = -(21GMem)/(21Re). Δ E = (11GMem)/(21Re), so x = 11, option (b).

An astronaut takes a ball of mass m from earth to space. He throws… | JEE Main 2024 (9 Apr, Shift 1)

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Satellite Energy & Weightlessness

An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5\text{ km}$ . From earth's surface to the orbit, the change in total mechanical energy of the ball is $x\frac{GM_e m}{21R_e}$ . The value of $x$ is (take $R_e = 6370\text{ km})$ :

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Answer: (b)

$h = 318.5$ km $= \frac{R_e}{20}$ , so $r = \frac{21R_e}{20}$ .
In orbit: $E_f = -\frac{GM_em}{2r} = -\frac{10GM_em}{21R_e}$ .
On the surface (at rest): $E_i = -\frac{GM_em}{R_e} = -\frac{21GM_em}{21R_e}$ .
$\Delta E = \frac{11GM_em}{21R_e}$ , so $x = 11$ , option (b).

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