NEET-XII-Physics
11: Gravitation
- #2The acceleration of the moon just before it strikes the earth in the previous question is
(a) 10 m s-2
(b) 0â‹…0027 m s-2
(c) 6â‹…4 m s-2
(d) 5â‹…0 m s-2digAnsr: cAns : (c) 6â‹…4 m s-2
According to the previous question, we have:
Radius of the moon, `` {R}_{m}=\frac{{R}_{e}}{4}=\frac{6400000}{4}=1600000\,\mathrm{\,m\,}``
So, when the Moon is just about to hit the surface of the Earth, its centre of mass is at a distance of (Re + Rm) from the centre of the Earth.
Acceleration of the Moon just before hitting the surface of the earth is given by
`` g\text{'}=\frac{GM}{({R}_{e}+{R}_{m}{)}^{2}}=\frac{GM}{{{R}_{e}}^{2}(1+{\displaystyle \frac{{R}_{m}}{{R}_{e}}}{)}^{2}}``
`` \Rightarrow g\text{'}=\frac{g}{(1+{\displaystyle \frac{{R}_{m}}{{R}_{e}}}{)}^{2}}=\frac{10}{(1+{\displaystyle \frac{1}{4}}{)}^{2}}=\frac{10\times 16}{25}``
`` \Rightarrow g\text{'}=6.4\,\mathrm{\,m\,}/{\,\mathrm{\,s\,}}^{2}``
Page No 224: