GRAVITY ON MARS
A robotic lander with an earth weight of 3430 N is sent to Mars, which has radius R_M = 3.39 × 10^6 m and mass m_M = 6.42 × 10^{23} kg (see Appendix F). Find the weight F_g of the lander on the Martian surface and the acceleration there due to gravity, g_M.
IDENTIFY and SET UP:
To find F_g we use Eq. (13.3), replacing m_E and R_E with m_M and R_M. We determine the lander mass m from the lander’s earth weight w and then find g_{\mathrm{M}} from F_g = mg_M.
w=F_{\mathrm{g}}=\frac{G m_{\mathrm{E}} m}{R_\mathrm{E}^{2}}EXECUTE:
The lander’s earth weight is w = mg, so
m=\frac{w}{g}=\frac{3430 \mathrm{~N}}{9.80 \mathrm{~m} / \mathrm{s}^{2}}=350 \mathrm{~kg}The mass is the same no matter where the lander is. From Eq. (13.3), the lander’s weight on Mars is
\begin{aligned}F_{\mathrm{g}} &=\frac{G m_{\mathrm{M}}^{m}}{R_{\mathrm{M}}^{2}} \\&=\frac{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(6.42 \times 10^{23} \mathrm{~kg}\right)(350 \mathrm{~kg})}{\left(3.39 \times10^{6} \mathrm{~m}\right)^{2}} \\&=1.30 \times 10^{3} \mathrm{~N}\end{aligned}The acceleration due to gravity on Mars is
g_{\mathrm{M}}=\frac{F_{\mathrm{g}}}{m}=\frac{1.30 \times 10^{3} \mathrm{~N}}{350 \mathrm{~kg}}=3.7 \mathrm{~m} / \mathrm{s}^{2}
EVALUATE: Even though Mars has just 11% of the earth’s mass \left(6.42 \times 10^{23} \mathrm{~kg} \text { versus } 5.97 \times 10^{24} \mathrm{~kg}\right), the acceleration due to
gravity g_M (and hence an object’s weight F_g) is roughly 40% as large as on earth. That’s because g_M is also inversely proportional to the square of the planet’s radius, and Mars has only 53% the radius of earth \left(3.39 \times 10^{6} \mathrm{~m} \text { versus } 6.37 \times 10^{6} \mathrm{~m}\right).
You can check our result for g_M by using Eq. (13.4), with appropriate replacements. Do you get the same answer?
g=\frac{G m_{\mathrm{E}}}{R_{\mathrm{E}}^{2}} (13.4)