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Gravitation question

2021 · 22 Jul · Shift 2 · Q53
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  5. /2021 · 22 Jul · Shift 2 · Q53

Gravitation question

2021 · 22 Jul · Shift 2 · Q53

JEE MainPhysicsGravitationMCQ+4 / −1
A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ‾\underline{\hspace{2cm}}​ s.
  1. A
    2Reg[(1+hRe)32−1]\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]g2Re​​​[(1+Re​h​)23​−1]
  2. B
    13Re2g[(1+hRe)32−1]{1 \over 3}\sqrt {{{{R_e}} \over {2g}}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]31​2gRe​​​[(1+Re​h​)23​−1]
  3. C
    Re2g[(1+hRe)32−1]\sqrt {{{{R_e}} \over {2g}}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]2gRe​​​[(1+Re​h​)23​−1]
  4. D
    132Reg[(1+hRe)32−1]{1 \over 3}\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]31​g2Re​​​[(1+Re​h​)23​−1]
View written solutionFree

Correct answer: D

  1. Given condition: velocity sufficient to carry the body to infinity

This means the body is projected with escape velocity from the surface of Earth.

So, v0=2GMRev_0=\sqrt{\frac{2GM}{R_e}}v0​=Re​2GM​​

Also, g=GMRe2⇒GM=gRe2g=\frac{GM}{R_e^2} \quad \Rightarrow \quad GM=gR_e^2g=Re2​GM​⇒GM=gRe2​

  1. Velocity at a height hhh above the surface

Let the distance from Earth's center at height hhh be r=Re+hr=R_e+hr=Re​+h

Using conservation of mechanical energy:

At launch from surface: E=12mv02−GMmReE=\frac12 m v_0^2-\frac{GMm}{R_e}E=21​mv02​−Re​GMm​

Since v0v_0v0​ is escape velocity, 12mv02=GMmRe\frac12 m v_0^2=\frac{GMm}{R_e}21​mv02​=Re​GMm​

Hence total energy is E=0E=0E=0

Therefore at distance rrr, 12mv2−GMmr=0\frac12 m v^2-\frac{GMm}{r}=021​mv2−rGMm​=0

So, 12mv2=GMmr\frac12 m v^2=\frac{GMm}{r}21​mv2=rGMm​ v=2GMrv=\sqrt{\frac{2GM}{r}}v=r2GM​​

Thus, drdt=2GMr\frac{dr}{dt}=\sqrt{\frac{2GM}{r}}dtdr​=r2GM​​

  1. Find time to reach height hhh

We need time from r=Retor=Re+hr=R_e \quad \text{to} \quad r=R_e+hr=Re​tor=Re​+h

So, dt=r2GM drdt=\sqrt{\frac{r}{2GM}}\,drdt=2GMr​​dr

Integrating,

=\frac{1}{\sqrt{2GM}}\int_{R_e}^{R_e+h} r^{1/2}dr$$ Now, $$\int r^{1/2}dr=\frac{2}{3}r^{3/2}$$ Therefore, $$t=\frac{1}{\sqrt{2GM}}\cdot \frac{2}{3}\left[(R_e+h)^{3/2}-R_e^{3/2}\right]$$ So, $$t=\frac{2}{3\sqrt{2GM}}\left[(R_e+h)^{3/2}-R_e^{3/2}\right]$$ --- 4. **Express in terms of $g$ and $R_e$** Using $$GM=gR_e^2$$ we get $$\sqrt{2GM}=\sqrt{2gR_e^2}=R_e\sqrt{2g}$$ Hence, $$t=\frac{2}{3R_e\sqrt{2g}}\left[(R_e+h)^{3/2}-R_e^{3/2}\right]$$ Factor out $R_e^{3/2}$: $$t=\frac{2}{3R_e\sqrt{2g}}\,R_e^{3/2}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]$$ Since $$\frac{2R_e^{3/2}}{3R_e\sqrt{2g}}=\frac{2\sqrt{R_e}}{3\sqrt{2g}}=\frac{1}{3}\sqrt{\frac{2R_e}{g}}$$ Thus, $$t=\frac{1}{3}\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]$$ --- 5. **Compare with options** This matches: $$\boxed{\frac{1}{3}\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]}$$ So the correct option is **D**. --- 6. **Verification with stored answer** Stored correct answer: **D** Derived answer: **D** They agree.
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