Determine the escape velocity from the Sun for an object at the Sun’s surface ( r = 7.0 x 10⁵ km , M = 2.0 x 10³⁰ kg).

Verified step by step guidance

1. Step 1: Recall the formula for escape velocity, which is derived from the conservation of energy. The escape velocity \( v_e \) is given by \( v_e = \sqrt{\frac{2GM}{r}} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \), \( M \) is the mass of the Sun, and \( r \) is the distance from the center of the Sun to the object.
2. Step 2: Convert the given radius \( r \) from kilometers to meters. Since \( 1 \ \text{km} = 1000 \ \text{m} \), multiply \( r = 7.0 \times 10^5 \ \text{km} \) by \( 1000 \) to get \( r = 7.0 \times 10^8 \ \text{m} \).
3. Step 3: Substitute the given values into the escape velocity formula. Use \( G = 6.674 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \), \( M = 2.0 \times 10^{30} \ \text{kg} \), and \( r = 7.0 \times 10^8 \ \text{m} \). The formula becomes \( v_e = \sqrt{\frac{2 \cdot (6.674 \times 10^{-11}) \cdot (2.0 \times 10^{30})}{7.0 \times 10^8}} \).
4. Step 4: Simplify the numerator \( 2 \cdot G \cdot M \) and the denominator \( r \) separately. Then divide the numerator by the denominator to find the value inside the square root.
5. Step 5: Take the square root of the result from Step 4 to find the escape velocity \( v_e \). Ensure the units are consistent throughout the calculation, and the final velocity will be in meters per second (m/s).

Key Concepts

Escape Velocity

Escape velocity is the minimum speed an object must reach to break free from the gravitational pull of a celestial body without any additional propulsion. It is derived from the balance between kinetic energy and gravitational potential energy, and is given by the formula v = √(2GM/r), where G is the gravitational constant, M is the mass of the body, and r is the radius from the center of the mass.

Gravitational Constant (G)

The gravitational constant, denoted as G, is a fundamental physical constant that quantifies the strength of gravitational attraction between two masses. Its value is approximately 6.674 × 10⁻¹¹ N(m/kg)². This constant is crucial in calculations involving gravitational forces and escape velocity, as it directly influences the gravitational potential energy of an object near a massive body.

Mass and Radius of the Sun

The mass and radius of the Sun are essential parameters in calculating escape velocity. The Sun's mass (approximately 2.0 x 10³⁰ kg) determines the strength of its gravitational field, while its radius (about 7.0 x 10⁵ km) defines the distance from its center at which the escape velocity is being calculated. These values are critical for applying the escape velocity formula accurately.