$$v_e = \sqrt{2GM_E/R_E}$$
Where:
v_e is the escape velocity
G is the gravitational constant (G ≈ 6.67430 x 10^-11 N·m²/kg²)
M_E is the mass of the Earth (M_E ≈ 5.972 x 10^24 kg)
R_E is the radius of the Earth (R_E ≈ 6.378 x 10^6 m)
Plugging in the values:
$$v_e = \sqrt{(2 x 6.67430 x 10^-11 N·m²/kg² x 5.972 x 10^24 kg)/(6.378 x 10^6 m)}$$
Calculating the result, we get:
$$v_e ≈ 11,180 m/s$$
So, the space shuttle needs to reach a speed of approximately 11,180 meters per second (about 25,000 miles per hour) to escape Earth's gravity and go into space.
