Question

the acceleration due to gravity on the moon is 1.6 m/s². what is the gravitational potential energy of a 1200-kg lander resting on top of a 350-m hill? 672000 j
how many times more gravitational potential energy would the same lander have on top of a 350-m hill on earth?
it would have times more gravitational potential energy.
4
6
10

Explanation:

Step1: Recall Gravitational PE Formula

Gravitational potential energy (PE) is given by \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.

Step2: Calculate PE on Moon

For the moon: \( m = 1200\,\text{kg} \), \( g_{\text{moon}} = 1.6\,\text{m/s}^2 \), \( h = 350\,\text{m} \).
\( PE_{\text{moon}} = 1200 \times 1.6 \times 350 \).
First, \( 1200 \times 1.6 = 1920 \). Then, \( 1920 \times 350 = 672000\,\text{J} \) (matches the given value).

Step3: Calculate PE on Earth

On Earth, \( g_{\text{earth}} = 9.8\,\text{m/s}^2 \) (approximate value, or note that \( g_{\text{earth}} \approx 6.125 \times g_{\text{moon}} \) since \( 9.8 / 1.6 \approx 6.125 \), but more accurately, \( g_{\text{earth}} \approx 9.8 \), \( g_{\text{moon}} = 1.6 \)).
\( PE_{\text{earth}} = 1200 \times 9.8 \times 350 \).

Step4: Find the Ratio

The ratio of \( PE_{\text{earth}} \) to \( PE_{\text{moon}} \) is \( \frac{PE_{\text{earth}}}{PE_{\text{moon}}} = \frac{m g_{\text{earth}} h}{m g_{\text{moon}} h} = \frac{g_{\text{earth}}}{g_{\text{moon}}} \).
Substitute \( g_{\text{earth}} = 9.8 \), \( g_{\text{moon}} = 1.6 \):
\( \frac{9.8}{1.6} \approx 6.125 \), but if using \( g_{\text{earth}} \approx 10 \) for approximation (or exact \( 9.8/1.6 = 6.125 \), but the options include 6? Wait, maybe the problem uses \( g_{\text{earth}} = 9.6 \) or a rounded value? Wait, \( 9.8 / 1.6 = 6.125 \), but the options have 6? Wait, maybe the intended \( g_{\text{earth}} = 9.8 \), \( g_{\text{moon}} = 1.6 \), so \( 9.8/1.6 \approx 6.125 \), but the closest option (if 6 is an option) or maybe the problem uses \( g_{\text{earth}} = 9.6 \)? Wait, no—wait, the first part's answer is 672000 J. Let's recalculate the ratio: \( 9.8 / 1.6 = 6.125 \), but maybe the problem expects using \( g_{\text{earth}} = 9.8 \) and \( g_{\text{moon}} = 1.6 \), so the ratio is \( 9.8 / 1.6 \approx 6.125 \), but the options given are 4, 6, 10. Wait, maybe a typo, but more accurately, \( g_{\text{earth}} \approx 9.8 \), \( g_{\text{moon}} = 1.6 \), so \( 9.8 / 1.6 = 6.125 \), which is approximately 6 (if 6 is an option) or maybe the problem uses \( g_{\text{earth}} = 9.6 \), then \( 9.6 / 1.6 = 6 \). Ah, that's likely! If \( g_{\text{earth}} \) is taken as 9.6 (a rounded value for simplicity), then \( 9.6 / 1.6 = 6 \). So the ratio is 6 (or approximately 6, since \( 9.8/1.6 \approx 6.125 \), close to 6).

Answer:

For the second part, the number of times more gravitational potential energy is approximately 6 (since \( \frac{g_{\text{earth}}}{g_{\text{moon}}} \approx \frac{9.6}{1.6} = 6 \) or \( \frac{9.8}{1.6} \approx 6.125 \), rounded to 6). So the answer is 6 (the option "6").