Answer :
To calculate the speed required to place a rock into orbit around asteroid 243 Ida, we'll need to rely on the principles of orbital mechanics. Specifically, we use the formula for the orbital speed of an object in orbit close to the surface of a spherical body.
Given data:
- Mass of asteroid ( [tex]\( M \)[/tex] ): [tex]\( 4.0 \times 10^{16} \, \text{kg} \)[/tex]
- Radius of asteroid ( [tex]\( R \)[/tex] ): [tex]\( 16 \, \text{km} \)[/tex] which is equivalent to [tex]\( 16000 \, \text{m} \)[/tex]
- Gravitational constant ( [tex]\( G \)[/tex] ): [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
The formula to calculate the orbital speed ( [tex]\( v \)[/tex] ) close to the surface of a celestial body is:
[tex]\[ v = \sqrt{\frac{G \cdot M}{R}} \][/tex]
Let’s break it down step-by-step:
1. Identify the known values:
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
- [tex]\( M = 4.0 \times 10^{16} \, \text{kg} \)[/tex]
- [tex]\( R = 16000 \, \text{m} \)[/tex]
2. Plug the known values into the formula:
[tex]\[ v = \sqrt{\frac{(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) \cdot (4.0 \times 10^{16} \, \text{kg})}{16000 \, \text{m}}} \][/tex]
3. Perform the calculations within the square root:
- First, compute the product of [tex]\( G \)[/tex] and [tex]\( M \)[/tex]:
[tex]\[ G \times M = (6.67430 \times 10^{-11}) \times (4.0 \times 10^{16}) = 2.66972 \times 10^6 \, \text{m}^3 \text{s}^{-2} \][/tex]
- Next, divide this product by [tex]\( R \)[/tex]:
[tex]\[ \frac{2.66972 \times 10^6 \, \text{m}^3 \text{s}^{-2}}{16000 \, \text{m}} = 166.85125 \, \text{m}^2 \text{s}^{-2} \][/tex]
4. Take the square root of the result:
[tex]\[ v = \sqrt{166.85125 \, \text{m}^2 \text{s}^{-2}} = 12.92 \, \text{m/s} \][/tex]
Therefore, the speed required to throw a rock into orbit around asteroid 243 Ida, near its surface, is approximately [tex]\( 12.92 \, \text{m/s} \)[/tex].
Given data:
- Mass of asteroid ( [tex]\( M \)[/tex] ): [tex]\( 4.0 \times 10^{16} \, \text{kg} \)[/tex]
- Radius of asteroid ( [tex]\( R \)[/tex] ): [tex]\( 16 \, \text{km} \)[/tex] which is equivalent to [tex]\( 16000 \, \text{m} \)[/tex]
- Gravitational constant ( [tex]\( G \)[/tex] ): [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
The formula to calculate the orbital speed ( [tex]\( v \)[/tex] ) close to the surface of a celestial body is:
[tex]\[ v = \sqrt{\frac{G \cdot M}{R}} \][/tex]
Let’s break it down step-by-step:
1. Identify the known values:
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]
- [tex]\( M = 4.0 \times 10^{16} \, \text{kg} \)[/tex]
- [tex]\( R = 16000 \, \text{m} \)[/tex]
2. Plug the known values into the formula:
[tex]\[ v = \sqrt{\frac{(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) \cdot (4.0 \times 10^{16} \, \text{kg})}{16000 \, \text{m}}} \][/tex]
3. Perform the calculations within the square root:
- First, compute the product of [tex]\( G \)[/tex] and [tex]\( M \)[/tex]:
[tex]\[ G \times M = (6.67430 \times 10^{-11}) \times (4.0 \times 10^{16}) = 2.66972 \times 10^6 \, \text{m}^3 \text{s}^{-2} \][/tex]
- Next, divide this product by [tex]\( R \)[/tex]:
[tex]\[ \frac{2.66972 \times 10^6 \, \text{m}^3 \text{s}^{-2}}{16000 \, \text{m}} = 166.85125 \, \text{m}^2 \text{s}^{-2} \][/tex]
4. Take the square root of the result:
[tex]\[ v = \sqrt{166.85125 \, \text{m}^2 \text{s}^{-2}} = 12.92 \, \text{m/s} \][/tex]
Therefore, the speed required to throw a rock into orbit around asteroid 243 Ida, near its surface, is approximately [tex]\( 12.92 \, \text{m/s} \)[/tex].
