Question 169: Gravitation

Question

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 ___________ s.

Accepted Answer

$${1 \over 2}m{v^2} - {{GMm} \over r} = 0 \Rightarrow v = \sqrt {{{2GM} \over r}} $$
$${{dr} \over {dt}} = \sqrt {{{2GM} \over r}} $$
$$ \Rightarrow \int\limits_{{R_e}}^{({R_e} + h)} {\sqrt r dr = \int\limits_0^t {\sqrt {2GM} dt} } $$
$$ \Rightarrow {2 \over 3}\left[ {{{({R_e} + h)}^{3/2}} - R_e^{3/2}} \right] = (t)\sqrt {2GM} $$
$$ \Rightarrow t = {1 \over 3}\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]$$

Answer

$$\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]$$

Answer

$${1 \over 3}\sqrt {{{{R_e}} \over {2g}}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]$$

Answer

$$\sqrt {{{{R_e}} \over {2g}}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]$$

Answer

$${1 \over 3}\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]$$

Gravitation

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 ___________ s.

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