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  • Suppose two planets (spherical in shape) of radii and but mass <img alt="\mathrm{M}$ and $9 \mathrm{M}" src="/latex-image/?%5Cmathrm%7BM%7D%24%20

Suppose two planets (spherical in shape) of radii R and 2 \mathrm{R}, but mass \mathrm{M}$ and $9 \mathrm{M} respectively have a centre to centre separation 8 \mathrm{R} as shown in the figure. A satellite of mass ' m 'is projected from the surface of the planet of mass 'M'directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is \sqrt{\frac{\mathrm{a}}{7} \frac{\mathrm{GM}}{\mathrm{R}}} then the value of 'a' is________.
[Given : The two planets are fixed in their position]
 

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best_answer

  

\rightarrowPt . P is a point where the net gravitational force is zero i.e force due to sphere of mass M and 9M are cancelling each other.

\rightarrow For the minimum speed of projection, Speed at point P must be zero.

F_{1}=F_{2}

\begin{aligned} \frac{G M M}{(R+x)^{2}} &=\frac{G(9 M)(M)}{(2 R+5 R-x)^{2}} \\ \frac{1}{R+x} &=\frac{3}{(7 R-x)} \\ 7 R-x &=3 R+3 x \\ 4 R &=4 x \\ x &=R \end{aligned}

By energy conservation,

\text {TE surface of mass m}= \text{TE}_{\text{P}}

\begin{gathered} -\frac{G M m}{R}-\frac{G(9 M) m}{7 R}+\frac{1}{2} m v^{2}=\frac{-G M m}{R+x}-\frac{G(9 m) m}{(7 R-x)}+1 \\ -\frac{16 G M m}{7 R}+\frac{1}{2} m v^{2}=-\frac{G M m}{2 R}-\frac{3}{2} \frac{G M m}{R} \\ -\frac{16}{7} \frac{G M m}{R}+\frac{1}{2} m v^{2}=-\frac{4 G M m}{2} \end{gathered}

                             \frac{1}{2}mv^2=\frac{-2 G M m}{R}+\frac{16}{7} \frac{G M m}{R}

                                                  \begin{aligned} \frac{1}{2} m v^{2} &=\frac{2 G M m}{7 R} \\ V &=\sqrt{\frac{4 G M}{7 R}} \\ \therefore a &=4 \end{aligned}

 

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vishal kumar

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