Question

Escape velocity from a planet of radius \(R\) and density \(\rho\) is given as \(10\,\text{km s}^{-1}\). Find the escape velocity from a planet of radius \(\dfrac{R}{10}\) and density \(\dfrac{\rho}{10}\).

• A. \(10\sqrt{100}\,\text{m s}^{-1}\)
• B. \(110\sqrt{10}\,\text{m s}^{-1}\)
• C. \(100\sqrt{10}\,\text{m s}^{-1}\)
• D. \(90\sqrt{10}\,\text{m s}^{-1}\)

Hint:

Remember:
\(v_e \propto R\sqrt{\rho}\)
For uniform density planets, scaling laws simplify calculations

Answer 1

The Correct Option is C

Step 1: Define intervals based on critical points.
The absolute value terms $|x+4|$ and $|x+2|$ change behavior at $x=-4$ and $x=-2$. We analyze the equation in three regions. 
Step 2: Case 1 ($x<-4$). 
Here, both $(x+4)$ and $(x+2)$ are negative. Equation: $x(-(x+4)) + 3(-(x+2)) + 10 = 0$ $\Rightarrow -x^2 - 4x - 3x - 6 + 10 = 0$ $\Rightarrow -x^2 - 7x + 4 = 0 \Rightarrow x^2 + 7x - 4 = 0$. Roots: $x = \frac{-7 \pm \sqrt{49+16}}{2} = \frac{-7 \pm \sqrt{65}}{2}$. Approximation ($\sqrt{65} \approx 8.1$): $x_1 \approx 0.55$ (Rejected, not $< -4$). $x_2 \approx -7.55$ (Accepted, $< -4$). 1 Solution found. 
Step 3: Case 2 ($-4 \le x<-2$). 
Here, $(x+4)$ is positive, $(x+2)$ is negative. Equation: $x(x+4) + 3(-(x+2)) + 10 = 0$ $\Rightarrow x^2 + 4x - 3x - 6 + 10 = 0 \Rightarrow x^2 + x + 4 = 0$. Discriminant $D = 1^2 - 16<0$. No real solution. 
Step 4: Case 3 ($x \ge -2$). 
Here, both are positive. Equation: $x(x+4) + 3(x+2) + 10 = 0$ $\Rightarrow x^2 + 7x + 16 = 0$. Discriminant $D = 49 - 64<0$. No real solution. Conclusion: Only 1 real solution exists.