The Asteroid433Eros has a mass of 7.2 x 1015 kg, and a radius of = 11 km. What is the escape velocity of this asteroid (remember to use r in meters)? What consequence would this have for someone trying to walk on this asteroid?
To calculate the escape velocity of an object, we can use the formula:
Escape velocity = √(2 * G * M / r)
Where:
- G is the gravitational constant (approximately 6.67 x 10^-11 N m²/kg²)
- M is the mass of the object (in kg)
- r is the radius of the object (in meters)
Given information:
Mass of asteroid (M) = 7.2 x 10^15 kg
Radius of asteroid (r) = 11 km = 11,000 meters
Using the formula, let's calculate the escape velocity of the asteroid:
Escape velocity = √(2 * 6.67 x 10^-11 N m²/kg² * 7.2 x 10^15 kg / 11000 meters)
Escape velocity = √(94.87 x 10^4 N m²/kg² * 654.55 x 10^15 kg / 11000 meters)
Escape velocity = √(62051.35 x 10^19 N kg / m)
Escape velocity = √(62051.35 x 10^19 N) = 7.87 x 10^9 m/s
Therefore, the escape velocity of the asteroid is approximately 7.87 x 10^9 m/s.
Now, let's consider the consequences of this escape velocity for someone trying to walk on the asteroid. The escape velocity represents the speed needed to overcome the gravitational pull and escape the asteroid's gravitational field. In simpler terms, it is the speed required to "launch" oneself off the surface of the asteroid and into space.
Since the escape velocity of this asteroid is extremely high (over 7.8 million meters per second), it would be practically impossible for an individual to walk or move on the asteroid's surface. Any attempt to walk would result in the person being launched off into space due to the asteroid's weak gravitational pull.
In summary, the escape velocity of 7.87 x 10^9 m/s on the asteroid Eros would make it impossible for someone to walk on its surface as they would be easily launched into space.