I just wanted to know the initial speed required to launch 1500 kg into 160 km orbit and that cargo should still have velocity of 7.8 km/s after reaching 160 km altitude. Does anyone know the proper formula?
This is what AI gives me:
To throw a 1 kg object into a 160 km orbit from sea level, we need to consider the total energy required to reach that orbit. This includes both the kinetic energy to reach orbital velocity and the potential energy to reach the altitude of 160 km. However, achieving orbit also means the object must attain the required horizontal velocity for a stable orbit, not just reach the altitude.
### Steps to Determine the Initial Speed:
1. **Gravitational Potential Energy at 160 km Altitude:**
The gravitational potential energy (\( U \)) at an altitude \( h \) is given by:
\[
U = mgh
\]
where \( m \) is the mass (1 kg), \( g \) is the acceleration due to gravity (9.81 m/s² at sea level), and \( h \) is the altitude (160,000 m).
2. **Orbital Velocity at 160 km Altitude:**
The orbital velocity (\( v_{orb} \)) at altitude \( h \) is given by:
\[
v_{orb} = \sqrt{\frac{GM}{R+h}}
\]
where \( G \) is the gravitational constant (\(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\)), \( M \) is the mass of the Earth (\(5.972 \times 10^{24} \, \text{kg}\)), and \( R \) is the Earth's radius (\(6.371 \times 10^6 \, \text{m}\)).
3. **Total Energy Required:**
The total energy required (\( E_{total} \)) to reach orbit includes both kinetic and potential energy:
\[
E_{total} = \frac{1}{2}mv_{orb}^2 + mgh
\]
### Calculations:
1. **Gravitational Potential Energy:**
\[
U = mgh = 1 \times 9.81 \times 160,000 = 1,569,600 \, \text{J}
\]
2. **Orbital Velocity:**
\[
v_{orb} = \sqrt{\frac{6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6 + 160,000}} \approx 7,784 \, \text{m/s}
\]
3. **Total Kinetic Energy at Orbital Velocity:**
\[
E_{kinetic} = \frac{1}{2}mv_{orb}^2 = \frac{1}{2} \times 1 \times (7,784)^2 \approx 30,300,000 \, \text{J}
\]
4. **Total Energy Required:**
\[
E_{total} = E_{kinetic} + U = 30,300,000 + 1,569,600 \approx 31,869,600 \, \text{J}
\]
To convert this total energy to the initial speed required at sea level, assuming no losses due to air drag or other factors (which in reality would be significant):
### Initial Speed Calculation:
1. **Initial Speed at Sea Level:**
\[
\frac{1}{2}mv_{initial}^2 = E_{total}
\]
\[
v_{initial} = \sqrt{\frac{2E_{total}}{m}} = \sqrt{\frac{2 \times 31,869,600}{1}} \approx 7,989 \, \text{m/s}
\]
This initial speed of approximately 7,989 m/s would be required in a theoretical, drag-free environment. However, considering atmospheric drag and other real-world factors, the actual required speed would be significantly higher. Real-world launches typically involve reaching much higher speeds initially to account for these losses and to achieve the desired orbit.
