For a rocket or spacecraft trying to leave the Earth without further propulsion, what is the approximate escape velocity from the Earth surface in km/s?

Introduction / Context:
This question checks your understanding of the concept of escape velocity in gravitation. Escape velocity is the minimum speed an object must have at a planet surface so that it can move away to an infinite distance without falling back, assuming no further thrust is applied and neglecting air resistance. Knowing the approximate value of escape velocity from the Earth is important in basic astronomy, space science and competitive exam physics.

Given Data / Assumptions:

• The object is a rocket or spacecraft launched from the surface of the Earth.
• No additional propulsion is applied after the launch speed is imparted.
• Air resistance is neglected to use the ideal theoretical formula.
• Standard values for the gravitational constant, mass of the Earth and radius of the Earth are assumed.
• We are asked for the approximate numeric value of escape velocity in km/s.

Concept / Approach:
Escape velocity is derived from energy conservation. To escape, a body must have enough initial kinetic energy to overcome the gravitational potential energy of the Earth. At the surface, the kinetic energy is (1/2) * m * v^2 and the gravitational potential energy relative to infinity is G * M * m / R, where G is the universal gravitational constant, M is the mass of the Earth, R is the radius of the Earth and m is the mass of the rocket. Setting the kinetic energy equal to the required potential energy gives the formula v_e = sqrt(2 * G * M / R). This formula shows that escape velocity does not depend on the mass of the rocket itself. Substituting Earth data gives an approximate value of about 11.2 km/s.

Step-by-Step Solution:
Step 1: Write the energy conservation condition for escape: initial kinetic energy = gravitational potential energy needed to reach infinite distance. Step 2: Use (1/2) * m * v_e^2 = G * M * m / R, where v_e is escape velocity. Step 3: Cancel the mass m of the rocket from both sides, showing escape velocity is independent of rocket mass. Step 4: Rearrange to get v_e^2 = 2 * G * M / R, so v_e = sqrt(2 * G * M / R). Step 5: Substitute typical values: G approximately 6.67 * 10^-11 N m^2/kg^2, M approximately 5.97 * 10^24 kg and R approximately 6.37 * 10^6 m. Step 6: Compute the result to get v_e about 11.2 * 10^3 m/s, which is 11.2 km/s. Step 7: Compare this with the options and identify 11.2 km/s as the correct choice.

Verification / Alternative check:
Many standard physics textbooks and reliable space science references quote the escape velocity from the Earth surface as approximately 11.2 km/s. Space missions typically use multistage rockets that provide continuous thrust and do not rely on achieving exactly this single impulse speed, but the value remains an important benchmark. Also, if you compute Earth orbital speed in low Earth orbit (around 7.9 km/s), the escape velocity is higher by a factor of about sqrt(2), which supports the numerical value near 11.2 km/s. These independent checks confirm that 11.2 km/s is the correct approximate escape speed.

Why Other Options Are Wrong:
10.2 km/s is lower than the required speed and would not allow a rocket to reach infinity without additional thrust; gravity would eventually pull it back. 12.2 km/s and 13.2 km/s are higher than the theoretical minimum and are not the standard quoted escape velocity value for Earth, even though speeds above 11.2 km/s would also allow escape. Only 11.2 km/s matches the widely accepted theoretical value derived from Earth mass and radius.

Common Pitfalls:
Students sometimes confuse escape velocity with the orbital velocity required for a circular orbit close to the Earth. Orbital velocity in low Earth orbit is about 7.9 km/s, which is significantly less than 11.2 km/s. Another misconception is to think that escape velocity depends on the mass of the rocket, but the derivation shows that mass cancels out. Remember that escape velocity depends only on the mass and radius of the planet or body from which the object is escaping, and for Earth this value is approximately 11.2 km/s.

Final Answer:
The approximate escape velocity from the Earth surface is 11.2 km/s.