Question:
Escape speed from the Earth is close to 11.2 km
s–1. On another planet whose radius is half of
the Earth’s radius and whose mass density is
four times that of the Earth, the escape speed
in km s–1 will be close to:
Escape speed from the Earth is close to 11.2 km
s–1. On another planet whose radius is half of
the Earth’s radius and whose mass density is
four times that of the Earth, the escape speed
in km s–1 will be close to:
Options:
- 11.2
- 15.8
- 5.6
- 7.9
- 11.2
- 15.8
- 5.6
- 7.9
Correct Answer: A) 11.2
Explanation:
Escape velocity v = sqrt(2GM/R) = sqrt(2G * p*(4/3)*pi*R^3 / R) = R * sqrt(8*pi*G*p/3).
Since radius is halved and density is 4 times, v_new = (R/2) * sqrt(4) * original_factor = original_velocity. Thus, it remains 11.2 km/s.
Escape velocity v = sqrt(2GM/R) = sqrt(2G * p*(4/3)*pi*R^3 / R) = R * sqrt(8*pi*G*p/3).
Since radius is halved and density is 4 times, v_new = (R/2) * sqrt(4) * original_factor = original_velocity. Thus, it remains 11.2 km/s.