Respuesta :
Answer: To find the speed of the International Space Station (ISS) in its orbit at an elevation of 350 km, we can use Newton's Law of Universal Gravitation. The formula for calculating the orbital speed of an object is:
v = √(G * M / r)
Where:
- v is the orbital speed
- G is the gravitational constant (approximately 6.67430 × 10^-11 N(m/kg)^2)
- M is the mass of the Earth (5.98 × 10^24 kg)
- r is the distance between the center of the Earth and the object's orbit (6378 km + 350 km)
Let's calculate the speed:
Step 1: Convert the units of Earth's radius and the ISS's elevation to meters.
Earth's radius = 6378 km = 6378000 m
ISS's elevation = 350 km = 350000 m
Step 2: Calculate the distance between the center of the Earth and the ISS's orbit.
r = Earth's radius + ISS's elevation = 6378000 m + 350000 m = 6728000 m
Step 3: Plug in the values into the formula and solve for v.
v = √(G * M / r)
v = √((6.67430 × 10^-11 N(m/kg)^2) * (5.98 × 10^24 kg) / 6728000 m)
After calculating this expression, we find that the speed of the International Space Station in its orbit at an elevation of 350 km is approximately 7.70 km/s.
Therefore, the correct answer is (a) 7.70 km/s.
Explanation:
Answer:
[tex]\sf \textsf{a. 7.70 km/s} [/tex]
Explanation:
Newton's law of universal gravitation is given by the equation:
[tex]\boxed{\boxed{\sf F = \dfrac{G \cdot m_1 \cdot m_2}{r^2} }}[/tex]
where:
- [tex]\sf F [/tex] is the gravitational force between two masses,
- [tex]\sf G [/tex] is the gravitational constant ([tex]\sf 6.67 \times 10^{-11} \, \textsf{Nm}^2/\textsf{kg}^2[/tex]),
- [tex]\sf m_1 [/tex] and [tex]\sf m_2 [/tex] are the masses of the two objects,
- [tex]\sf r [/tex] is the separation between the centers of the masses or radius.
The centripetal force required to keep an object in circular motion is given by:
[tex]\sf F = \dfrac{m \cdot v^2}{r} [/tex]
where:
- [tex]\sf m [/tex] is the mass of the object,
- [tex]\sf v [/tex] is its velocity,
- [tex]\sf r [/tex] is the radius of the circular path.
In the case of the International Space Station (ISS) in orbit around Earth, the gravitational force provides the centripetal force:
[tex]\sf \dfrac{G \cdot m_{\textsf{Earth}} \cdot m_{\textsf{ISS}}}{r^2} = \dfrac{m_{\textsf{ISS}} \cdot v^2}{r} [/tex]
Now, we can solve for [tex]\sf v [/tex] (the orbital speed of the ISS).
Given:
- [tex]\sf r [/tex] (distance from Earth's center to the ISS) is [tex]\sf 350 \, \textsf{km} + 6378 \, \textsf{km}[/tex],
- [tex]\sf m_{\textsf{Earth}} = 5.98 \times 10^{24} \, \textsf{kg} [/tex],
- [tex]\sf m_{\textsf{ISS}} = 304,000 \, \textsf{kg} [/tex].
Now, we have:
[tex]\sf \dfrac{G \cdot m_{\textsf{Earth}} \cdot m_{\textsf{ISS}}}{(r_{\textsf{ISS}})^2} = \dfrac{m_{\textsf{ISS}} \cdot v^2}{r_{\textsf{ISS}}} [/tex]
[tex]\sf v = \sqrt{\dfrac{G \cdot m_{\textsf{Earth}}}{r_{\textsf{ISS}}}} [/tex]
Substitute the value:
[tex]\sf v = \sqrt{\dfrac{6.67 \times 10^{-11} \, \textsf{Nm}^2/\textsf{kg}^2 \cdot 5.98 \times 10^{24} \, \textsf{kg}}{r_{\textsf{ISS}}}} [/tex]
[tex]\sf v = \sqrt{\dfrac{6.67 \times 10^{-11} \, \textsf{Nm}^2/\textsf{kg}^2 \cdot 5.98 \times 10^{24} \, \textsf{kg}}{(6378 \, \textsf{km} + 350 \, \textsf{km}) \cdot 1000 \, \textsf{m/km}}} [/tex]
After calculating this expression, we get:
[tex]\sf v = \sqrt{\dfrac{ 3.98866\times 10^{14}}{6728000 }}[/tex]
[tex]\sf v = \sqrt{59284482.76 }[/tex]
[tex]\sf v = 7699.641729 \, \textsf{m/s} [/tex]
[tex]\sf v = 7.699641729 \, \textsf{km/s} [/tex]
[tex]\sf v \approx 7.7 \, \textsf{km/s (in nearest tenth)} [/tex]
So, the correct answer is:
[tex]\sf \textsf{a. 7.70 km/s} [/tex]
