Answer :
Let's solve the problem step by step.
First, we'll calculate the gravitational acceleration on the surface of the Earth as it is now, using the formula for gravitational acceleration:
[tex]\[ g = \frac{G \cdot M}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant, which is [tex]\( 6.67430 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex],
- [tex]\( M \)[/tex] is the mass of the Earth, which is [tex]\( 5.972 \times 10^{24} \ \text{kg} \)[/tex],
- [tex]\( r \)[/tex] is the radius of the Earth.
The average radius of the Earth is about 6371 kilometers. We need to convert this to meters:
[tex]\[ r_{\text{Earth}} = 6371 \ \text{km} \times 1000 \ \text{m/km} = 6371000 \ \text{m} \][/tex]
So, the gravitational acceleration on Earth is:
[tex]\[ g_{\text{Earth}} = \frac{G \cdot M}{r_{\text{Earth}}^2} = \frac{6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{(6371000)^2} \][/tex]
Next, let's calculate the gravitational acceleration on a hypothetical Earth compressed to the size of the Moon. The diameter of the Moon is 3474 km, so the radius is:
[tex]\[ r_{\text{Moon}} = \frac{3474 \ \text{km}}{2} \times 1000 \ \text{m/km} = 1737000 \ \text{m} \][/tex]
Using the same formula for gravitational acceleration, we have:
[tex]\[ g_{\text{compressed}} = \frac{G \cdot M}{r_{\text{Moon}}^2} = \frac{6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{(1737000)^2} \][/tex]
Now, to find how many times the gravitational acceleration on the compressed Earth is larger than that on the real Earth, we take the ratio of [tex]\( g_{\text{compressed}} \)[/tex] to [tex]\( g_{\text{Earth}} \)[/tex]:
[tex]\[ \text{Change in gravity} = \frac{g_{\text{compressed}}}{g_{\text{Earth}}} = \frac{\frac{G \cdot M}{r_{\text{Moon}}^2}}{\frac{G \cdot M}{r_{\text{Earth}}^2}} \][/tex]
Notice that the mass [tex]\( M \)[/tex] and the gravitational constant [tex]\( G \)[/tex] are the same for both and cancel out in the ratio, so we have:
[tex]\[ \text{Change in gravity} = \frac{r_{\text{Earth}}^2}{r_{\text{Moon}}^2} \][/tex]
Plug in the values and simplify:
[tex]\[ \text{Change in gravity} = \frac{(6371000)^2}{(1737000)^2} \approx \frac{(6371)^2}{(1737)^2} \approx \frac{40581641}{3018369} \approx 13.45 \][/tex]
So, the acceleration due to gravity of the hypothetical compressed Earth is approximately 13.45 times the acceleration due to gravity of the real Earth. Note that this result may slightly vary due to rounding during the calculation.
First, we'll calculate the gravitational acceleration on the surface of the Earth as it is now, using the formula for gravitational acceleration:
[tex]\[ g = \frac{G \cdot M}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant, which is [tex]\( 6.67430 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex],
- [tex]\( M \)[/tex] is the mass of the Earth, which is [tex]\( 5.972 \times 10^{24} \ \text{kg} \)[/tex],
- [tex]\( r \)[/tex] is the radius of the Earth.
The average radius of the Earth is about 6371 kilometers. We need to convert this to meters:
[tex]\[ r_{\text{Earth}} = 6371 \ \text{km} \times 1000 \ \text{m/km} = 6371000 \ \text{m} \][/tex]
So, the gravitational acceleration on Earth is:
[tex]\[ g_{\text{Earth}} = \frac{G \cdot M}{r_{\text{Earth}}^2} = \frac{6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{(6371000)^2} \][/tex]
Next, let's calculate the gravitational acceleration on a hypothetical Earth compressed to the size of the Moon. The diameter of the Moon is 3474 km, so the radius is:
[tex]\[ r_{\text{Moon}} = \frac{3474 \ \text{km}}{2} \times 1000 \ \text{m/km} = 1737000 \ \text{m} \][/tex]
Using the same formula for gravitational acceleration, we have:
[tex]\[ g_{\text{compressed}} = \frac{G \cdot M}{r_{\text{Moon}}^2} = \frac{6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{(1737000)^2} \][/tex]
Now, to find how many times the gravitational acceleration on the compressed Earth is larger than that on the real Earth, we take the ratio of [tex]\( g_{\text{compressed}} \)[/tex] to [tex]\( g_{\text{Earth}} \)[/tex]:
[tex]\[ \text{Change in gravity} = \frac{g_{\text{compressed}}}{g_{\text{Earth}}} = \frac{\frac{G \cdot M}{r_{\text{Moon}}^2}}{\frac{G \cdot M}{r_{\text{Earth}}^2}} \][/tex]
Notice that the mass [tex]\( M \)[/tex] and the gravitational constant [tex]\( G \)[/tex] are the same for both and cancel out in the ratio, so we have:
[tex]\[ \text{Change in gravity} = \frac{r_{\text{Earth}}^2}{r_{\text{Moon}}^2} \][/tex]
Plug in the values and simplify:
[tex]\[ \text{Change in gravity} = \frac{(6371000)^2}{(1737000)^2} \approx \frac{(6371)^2}{(1737)^2} \approx \frac{40581641}{3018369} \approx 13.45 \][/tex]
So, the acceleration due to gravity of the hypothetical compressed Earth is approximately 13.45 times the acceleration due to gravity of the real Earth. Note that this result may slightly vary due to rounding during the calculation.
