Source: High school physics (Chinese)
Problem Sets:
Problem
The average diameter of Mars is $6.79 \times 10^3$ km, and the average diameter of Earth is $1.28 \times 10^4$ km. The mass of Mars is 0.108 times the mass of Earth. Let $g_E = 9.8$ m/s$^2$.
- What is the ratio of the average density of Mars to that of Earth?
- What is the gravitational acceleration on Mars?
- What is the escape velocity for an object on Mars?
[Q1] 0.724 [Q2] $g_M = 3.76$ m/s$^2$ [Q3] $v_{e,M} = 5.05 \times 10^3$ m/s
Let subscripts M denote Mars and E denote Earth. Given data: $D_M = 6.79 \times 10^3$ km, $D_E = 1.28 \times 10^4$ km, $M_M = 0.108 M_E$. This gives radius ratio $\frac{R_E}{R_M} = \frac{D_E}{D_M} = \frac{1.28 \times 10^4}{6.79 \times 10^3} \approx 1.885$.
[Q1] Density is $\rho = M/V = M/(\frac{4}{3}\pi R^3)$. The ratio is:
$$\frac{\rho_M}{\rho_E} = \frac{M_M}{M_E} \left(\frac{R_E}{R_M}\right)^3 = (0.108)(1.885)^3 \approx 0.724$$[Q2] Gravitational acceleration is $g = GM/R^2$. The ratio is:
$$\frac{g_M}{g_E} = \frac{M_M}{M_E} \left(\frac{R_E}{R_M}\right)^2 = (0.108)(1.885)^2 \approx 0.384$$ $$g_M = 0.384 g_E = 0.384(9.8 \text{ m/s}^2) \approx 3.76 \text{ m/s}^2$$[Q3] Escape velocity is $v_e = \sqrt{2gR}$.
$$v_{e,M} = \sqrt{2 g_M R_M}$$ $R_M = D_M/2 = (6.79 \times 10^3 \text{ km})/2 = 3.395 \times 10^6$ m. $$v_{e,M} = \sqrt{2(3.76 \text{ m/s}^2)(3.395 \times 10^6 \text{ m})} \approx 5.05 \times 10^3 \text{ m/s}$$