If a planet of mass 6.4 times 10^{23} kg can | vedclass

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(B) The escape velocity $v_e$ of a planet is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$ Squaring both sides,we get: $v_e^2 = \frac{2GM}{R}$ Re

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$13.2$

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(B) The escape velocity $v_e$ of a planet is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$ Squaring both sides,we get: $v_e^2 = \frac{2GM}{R}$ Rearranging the formula to solve for the radius $R$: $R = \frac{2GM}{v_e^2}$ Given values: $M = 6.4 	imes 10^{23} \ kg$ $v_e = 8 	imes 10^4 \ m/s$ $G = 6.67 	imes 10^{-11} \ N \cdot m^2/kg^2$ Substituting these values into the equation: $R = \frac{2 	imes 6.67 	imes 10^{-11} 	imes 6.4 	imes 10^{23}}{(8 	imes 10^4)^2}$ $R = \frac{85.376 	imes 10^{12}}{64 	imes 10^8}$ $R = 1.334 	imes 10^4 \ m \approx 13.3 	imes 10^3 \ m = 13.3 \ km$ Considering the approximation used in the provided options,the closest value is $13.2 \ km$. Thus,option $B$ is correct.

If a planet of mass $6.4 \times 10^{23} \ kg$ can be compressed into a sphere such that the escape velocity from its surface is $8 \times 10^4 \ m/s$,then what should be the radius of the sphere (in $km$)? (Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

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