Explanation
Given that,
Acceleration $$a=2.45\,m/{{s}^{2}}$$
Mass of earth $$M=6\times {{10}^{24}}\,kg$$
Radius of earth $$R=6\times {{10}^{6}}\,m$$
Gravitational constant $$G=6.67\times {{10}^{-11}}\,N{{m}^{2}}/kg$$
Now centripetal acceleration is
$$ {{a}_{c}}=\dfrac{{{v}^{2}}}{R} $$
$$ {{a}_{c}}=\dfrac{{{\left( \sqrt{\dfrac{GM}{R}} \right)}^{2}}}{R} $$
$$ {{a}_{c}}=\dfrac{GM}{{{R}^{2}}} $$
$$ {{a}_{c}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}} $$
Now, put the values
$$ 2.45=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{\left( 6\times {{10}^{6}}+h \right)}^{2}}} $$
$$ {{\left( 6\times {{10}^{6}}+h \right)}^{2}}=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{2.45} $$
$$ {{\left( 6\times {{10}^{6}}+h \right)}^{2}}=\dfrac{40.02\times {{10}^{13}}}{2.45} $$
$$ {{\left( 6\times {{10}^{6}}+h \right)}^{2}}=1.6334\times {{10}^{14}} $$
$$ 6\times {{10}^{6}}+h=\sqrt{1.6334\times {{10}^{14}}} $$
$$ h=1.278\times {{10}^{7}}-6\times {{10}^{6}} $$
$$ h=\left( 12.78-6 \right)\times {{10}^{6}} $$
$$ h=6.78\times {{10}^{6}} $$
$$ h=6.8\times {{10}^{6}}\,m $$
So, $$h\cong R$$
Hence, the height of spaceship is $$R$$
If t1 and t2 are the time when the body is at the same height Then $$h = \dfrac{1}{2} \times g\times {t}_1\times {t}_2$$ $$h = \dfrac{1}{2} \times g\times {2}\times {10}$$h = 10 g
So that the correct option is D
$$ g=9.8\,m/{{s}^{2}} $$
$$ h=480\,km $$
$$ R=6400\,km $$
We know that,
$$ g'=g\left( 1-\dfrac{2h}{R} \right) $$
$$ g'=9.8\left( 1-\dfrac{2\times 480}{6400} \right) $$
$$ g'=8.33\,m/{{s}^{2}} $$
Hence, the value of g’ is $$8.33\,m/{{s}^{2}}$$.
Net force on the rocket is zero. i.e. force due to the earth and force due to moon is same but in opposite direction.$$Fe=Fm$$$$\dfrac{GMe Mr}{re^2}=\dfrac{GMmMr}{rm^2}$$$$r$$ is the total distance$$r= re+ rm$$$$Me=81 Mm$$$$\dfrac{a 81 Mm Mr}{re^2}=\dfrac{G M m Mr}{rm^2}$$$$r_m=\dfrac{r_e}{g}$$$$g rm +rm =r \Rightarrow 10 r_m=r$$$$\Rightarrow r_m=\dfrac{r}{10}$$
Weight of body $$W=144\,N$$
Height $$h=3R$$
Radius of earth $$=R$$
Acceleration due to gravity at the surface of the earth $$=g$$
Acceleration due to gravity at height $$g'=3R$$
Now,
$$ \dfrac{g}{g'}={{\left( \dfrac{R}{R+h} \right)}^{2}} $$
$$ \dfrac{g}{g'}={{\left( \dfrac{R}{R+3R} \right)}^{2}} $$
$$ \dfrac{g}{g'}=\dfrac{1}{16} $$
Now, weight at the height
$$ \dfrac{w'}{w}=\dfrac{mg'}{mg} $$
$$ \dfrac{w'}{w}=\dfrac{1}{16} $$
$$ w'=\dfrac{w}{16} $$
$$ w'=\dfrac{144}{16} $$
$$ w'=9\,N $$
Hence, the weight at the height is $$9\ N$$
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