If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is three-fifths the escape speed?
I keep getting the answer 2.56x10^6 m after solving it a few different ways. Help would be great.
Its energy at launch is PE at surface of earth.
1/2 mv^2=GMe*M/Re
so when is GPE= (2/5)^2*GMeM/Re ?
or GMeM/(Re+H)=4/25*GMeM/Re
4(Re+h)=25Re
or 4h=21Re
h= 21Re/4 or near 5.25Re
or h= about 6.5*5.2E6 meters which is not your answer. check my work.
To solve this problem, we can use conservation of mechanical energy. The total mechanical energy of the projectile remains constant throughout its motion.
The escape speed is the minimum speed required for an object to escape the gravitational pull of the Earth. At the escape speed, the total mechanical energy of the projectile is zero because it has just enough kinetic energy to overcome the gravitational potential energy.
We can write the equation for the total mechanical energy as the sum of kinetic energy and gravitational potential energy:
1/2 mv^2 - GMm/r = 0
Where:
m = mass of the projectile
v = speed of the projectile
G = gravitational constant
M = mass of the Earth
r = distance between the center of the Earth and the projectile
Now, let's solve for the height above the Earth's surface when the projectile's speed is three-fifths the escape speed.
1. Set up the equation for the total mechanical energy:
1/2 mv^2 - GMm/r = 0
2. Substitute v = (3/5)vescape, where vescape is the escape speed, into the equation:
1/2 m((3/5)vescape)^2 - GMm/r = 0
3. Simplify the equation:
9/50 mvescape^2 - GMm/r = 0
4. Rearrange the equation to solve for r:
r = GMm / (9/50 mvescape^2)
5. Cancel out the mass:
r = G / (9/50 vespace^2)
6. Substitute the values of G and vescape:
r = (6.674 x 10^-11 m^3/kg/s^2) / (9/50 (11,200 m/s)^2)
7. Calculate the value of r:
r = (6.674 x 10^-11 m^3/kg/s^2) / (9/50 (11,200 m/s)^2)
= (6.674 x 10^-11 m^3/kg/s^2) / (9/50 (125,440,000 m^2/s^2))
= 2.560 x 10^6 meters
Therefore, the projectile is approximately 2.56 x 10^6 meters above the Earth's surface when its speed is three-fifths the escape speed.