If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how far from the Earth's center is it when its speed is 1/9 of the escape speed?
v=sqrt(2•G•M•R)
v1 = v/9 = sqrt(2•G•M•R)/9,
r =81•R = 81•6.4•10^6 m =5.18•10^8 m
To solve this problem, we can use the concept of conservation of mechanical energy. The mechanical energy of the projectile is conserved throughout its vertical motion.
1. The escape speed, v_esc, is the minimum speed required for a projectile to escape the gravitational pull of Earth. It is given by the equation:
v_esc = √(2GM/R),
where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
2. Let's denote the initial speed of the projectile (when launched vertically) as v_0, and the distance of the projectile from the Earth's center as r.
3. The mechanical energy of the projectile can be expressed as the sum of the kinetic energy and the gravitational potential energy:
E = (1/2)mv^2 - GMm/r,
where m is the mass of the projectile.
4. At the escape speed, the mechanical energy is zero:
E = (1/2)mv_esc^2 - GMm/r = 0.
5. When the speed of the projectile is 1/9 of the escape speed, the mechanical energy is given by:
E' = (1/2)mv^2 - GMm/r' = (1/18)mv_esc^2 - GMm/r' = 0,
where r' is the distance of the projectile from the Earth's center when its speed is 1/9 of the escape speed.
6. Rearranging the equation in Step 4 and Step 5, we get:
v_esc^2 = 2GM/r, (Equation 1)
(1/18)mv_esc^2 = GMm/r'. (Equation 2)
7. Divide Equation 2 by Equation 1 and simplify:
(1/18) = r'/r.
8. Rearrange the equation to solve for r':
r' = (1/18)r.
So, when the speed of the projectile is 1/9 of the escape speed, it is located at a distance of 1/18 times the original distance from the Earth's center.
To find the distance from the Earth's center when the speed of a projectile is 1/9 of the escape speed, we need to understand the concepts of escape speed and kinetic energy.
1. Escape Speed:
Escape speed is the minimum speed required for a projectile to completely escape the gravitational field of a celestial body like the Earth. Mathematically, the escape speed can be calculated using the formula:
v_escape = √(2 * G * M / R)
where v_escape is the escape speed, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the celestial body (in this case, Earth), and R is the radius of the celestial body (in this case, Earth's radius).
2. Kinetic Energy:
The kinetic energy of a projectile is given by the formula:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the projectile, and v is the velocity of the projectile.
To determine how far from Earth's center the projectile is when its speed is 1/9 of the escape speed, we can equate the kinetic energy of the projectile at that point to 1/2 of its kinetic energy at the escape speed.
Let's walk through the steps to solve the problem:
Step 1: Find the escape speed.
- Use the formula v_escape = √(2 * G * M / R) to calculate the escape speed of the projectile launched from Earth.
Step 2: Calculate the kinetic energy at the escape speed.
- Use the formula KE = (1/2) * m * v^2 to find the kinetic energy of the projectile at the escape speed.
Step 3: Find the speed when the kinetic energy is 1/2 of the escape speed.
- Multiply the kinetic energy at the escape speed by 1/9 to find the kinetic energy at 1/9 of the escape speed.
- Rearrange the kinetic energy formula to solve for v.
- Substitute the calculated value of kinetic energy at 1/9 of the escape speed to find the velocity of the projectile.
Step 4: Find the distance from Earth's center.
- Since the projectile is launched vertically, its speed will decrease as it moves away from Earth's surface.
- Use the concept of conservation of mechanical energy to relate the kinetic energy of the projectile at different distances from Earth's center.
- Equate the kinetic energy at the escape speed to the kinetic energy at 1/9 of the escape speed and solve for the distance from Earth's center.
By following these steps, you can calculate the distance from the Earth's center when the projectile's speed is 1/9 of the escape speed.