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 \fracGM_e m21 R_e . The value of x is (take R_e=6370 ~km) :

Gravitation PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q78 with Solution

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Correct answer: B

If the ball is at rest on the Earth's surface, its total mechanical energy is just gravitational potential energy:

$E_i = -\frac{GM_e m}{R_e}$

For a satellite in a circular orbit of radius $r$ , total mechanical energy is:

$E_f = -\frac{GM_e m}{2r}$

Here altitude is $h = 318.5\,\text{km}$ and $R_e = 6370\,\text{km}$ , so

$h = \frac{6370}{20} = 0.05R_e$

Thus orbit radius:

$r = R_e + h = R_e + 0.05R_e = 1.05R_e = \frac{21}{20}R_e$

So,

= -\frac{GM_e m}{21R_e/10} = -\frac{10GM_e m}{21R_e}$$ 3. **Change in total mechanical energy** $$\Delta E = E_f - E_i$$ $$\Delta E = -\frac{10GM_e m}{21R_e} - \left(-\frac{GM_e m}{R_e}\right)$$ $$\Delta E = -\frac{10GM_e m}{21R_e} + \frac{GM_e m}{R_e}$$ Write the second term with denominator $21R_e$: $$\frac{GM_e m}{R_e} = \frac{21GM_e m}{21R_e}$$ Hence, $$\Delta E = \frac{21GM_e m}{21R_e} - \frac{10GM_e m}{21R_e} = \frac{11GM_e m}{21R_e}$$ Comparing with $$x\frac{GM_e m}{21R_e}$$ we get $$x = 11$$ 4. **Option check** - A: $12$ ❌ - B: $11$ ✅ - C: $9$ ❌ - D: $10$ ❌ Therefore, the correct answer is **Option B**.