The speed with which the rock crashes into a planet is [tex]v=1.345 m/s[/tex].
Given,
The mass of planet 1 is 3M
The radius of planet 1 is R
The mass of planet 2 is 4M
The radius of planet 2 is 2R
Distance between the two planets; r = 8R
The value of R is; [tex]R=5.8[/tex] × [tex]10^{6} m[/tex]
The value of M is; [tex]M=7.3[/tex] × [tex]10^{23} kg[/tex]
Using Newton's universal gravitational law, the gravitational force on the rock (4R)etween the two planets is computed as follows:
[tex]F=\frac{G(3M)(4M)}{(4R)^{2} }[/tex]
[tex]F=\frac{G(12M^{2} )}{16R^{2} }[/tex]
Where,
G = Universal Gravitational Constant = [tex]6.6743 * 10^{-11} \frac{m^{3} }{kgs^{2} }[/tex]
On solving,
[tex]F=\frac{(6.67*10^{-11} )*12*(7.3*10^{23} )^{2} }{16(5.8*10^{6}) }[/tex]
[tex]F=7.925[/tex] × [tex]10^{23} N[/tex]
Now, the acceleration of rock is calculated as,
We know that,
Kinetic Energy; [tex]KE=\frac{1}{2}mv^{2}[/tex] and
Potential Energy; [tex]PE=\frac{GMm}{R}[/tex]
Applying the law of conservation of energy,
[tex]\frac{3GMm}{4R} -\frac{4GMm}{4R} = \frac{1}{2} mv^{2}+\frac{3GMm}{6R} -\frac{4GMm}{2R}[/tex]
[tex]-\frac{GMm}{4R}=\frac{1}{2}mv^{2} -\frac{3GMm}{2R}[/tex]
[tex]\frac{1}{2} mv^{2} = \frac{5GMm}{4R}[/tex]
[tex]v^{2}=\frac{5GM }{2R}[/tex]
[tex]v =\sqrt{\frac{5GM}{2R} }[/tex]
Putting the values of G, M and R we get
[tex]v=\sqrt{\frac{5*6.6743*10^{-11} *7.3*10^{23} }{2*5.8*10^{6} } }[/tex]
[tex]v=1.345 m/s[/tex]
The speed with which the rock crashes into a planet is [tex]v=1.345 m/s[/tex].
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