calculate the escape velocity of a satellite from the earth's gravitational field,if the radius of the earth is 6.4×10^6m
as you know,
v^2 = 2GM/r
Now just look up G and Mearth and crank it out.
Answer
To calculate the escape velocity of a satellite from Earth's gravitational field, you can use the following formula:
Escape velocity (V) = √(2 * gravitational constant (G) * mass of Earth (M) / radius of Earth (r))
Here, the gravitational constant (G) can be approximated as 6.674 × 10^-11 Nm^2/kg^2, and the mass of Earth (M) is approximately 5.972 × 10^24 kg.
Given that the radius of Earth (r) is 6.4 × 10^6 m, we can substitute the values into the formula:
V = √(2 * 6.674 × 10^-11 Nm^2/kg^2 * 5.972 × 10^24 kg / 6.4 × 10^6 m)
Simplifying:
V = √(2 * (6.674 × 5.972) / 6.4) × 10^18
V = √(80.158) × 10^18
V ≈ 2.37 × 10^4 m/s
So, the escape velocity of a satellite from Earth's gravitational field is approximately 2.37 × 10^4 m/s.
To calculate the escape velocity of a satellite from the Earth's gravitational field, we can use the formula:
Escape velocity (Ve) = √(2 * g * R)
Where:
Ve = Escape velocity
g = Acceleration due to gravity
R = Radius of the Earth
Given that the radius of the Earth (R) is 6.4 × 10^6 meters, we need to determine the value of acceleration due to gravity (g). The value of g can be calculated using the formula:
g = G * (M / R^2)
Where:
G = Universal gravitational constant (approximately 6.67430 × 10^-11 Nm^2/kg^2)
M = Mass of the Earth (approximately 5.972 × 10^24 kg)
Now, let's calculate the value of g:
g = (6.67430 × 10^-11 Nm^2/kg^2) * (5.972 × 10^24 kg) / (6.4 × 10^6 m)^2
Simplifying the above equation:
g ≈ 9.81 m/s^2
Now, substitute the value of g and R into the escape velocity formula:
Ve = √(2 * 9.81 m/s^2 * 6.4 × 10^6 m)
Calculating the above equation:
Ve ≈ √(2 * 9.81 m/s^2 * 6.4 × 10^6 m) ≈ √(125,952,000 m^2/s^2) ≈ 11,200 m/s
Therefore, the escape velocity of a satellite from the Earth's gravitational field, given a radius of the Earth as 6.4 × 10^6 meters, is approximately 11,200 m/s.