The international space center is 1500km above the earths surface. Determine the escape velocity of a rocket launched into space at this altitude if the mass of the earth is 6*10^24kg and the radius is 6.4*10^6m.
(1/2) m v^2 = difference in potential energy between rocket at infinity and at r =( 6.4*10^6 + 1.5*10^6) meters = G m M/r
where G = 6.67*10^-11
v^2 = 2 (6.67*10^-11)(6*10^24)/7.9*10^6
To determine the escape velocity of a rocket launched into space at a given altitude, we need to use the equation for escape velocity:
ve = √(2GM/r)
where ve is the escape velocity, G is the gravitational constant, M is the mass of the celestial body (in this case, Earth), and r is the distance from the center of the celestial body.
First, we need to convert the altitude of the International Space Center (1500 km) into meters:
Altitude = 1500 km = 1500 × 1000 = 1,500,000 m
Next, we need to calculate the total distance from Earth's center at this altitude:
r = radius of the Earth + altitude
= 6.4 × 10^6 m + 1,500,000 m
= 7.9 × 10^6 m
Now we can calculate the escape velocity using the equation above. We'll also need the value for the gravitational constant, which is approximately 6.67430 × 10^(-11) m^3/(kg s^2):
G = 6.67430 × 10^(-11) m^3/(kg s^2)
Substituting the values into the equation, we get:
ve = √(2 × G × M / r)
= √(2 × 6.67430 × 10^(-11) × 6 × 10^24 / 7.9 × 10^6)
Now we can calculate the escape velocity by evaluating the expression inside the square root:
ve = √((1.33486 × 10^(14)) / (7.9 × 10^6))
≈ 11,181.74 m/s
Therefore, the escape velocity of a rocket launched into space at an altitude of 1500 km above Earth's surface is approximately 11,181.74 m/s.