- #1
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- Homework Statement
- Jack can jump upwards a distance 1.4 meters when he is on the surface of the Earth. What is his initial velocity when he jumps?
Most of the time, however, Jack is in space. He is an asteroid miner: he looks for asteroids made out of useful metals. His job involves landing on asteroids and taking samples for analysis.
Jack knows that when he is on an asteroid, the local gravitational force can be very small. He must be careful -- if he stands up too fast or steps too hard, he could launch himself into space and never return.
One day, Joe lands on a small spherical asteroid of iron which has a radius of R=280 meters. What is the escape velocity from this body?
"Nope," says Joe, "it would be too dangerous to walk around on that asteroid. I might accidentally push myself away and never return."
What is the minimum radius a spherical asteroid of pure iron could have and still be "safe" for Jack? That is, how big must an iron asteroid be in order to prevent Jack from leaping off it by accident?
- Relevant Equations
- F= -G(m1)(m2)/(radius)^2
G=(6.67*10^-11)
Homework Statement: Jack can jump upwards a distance 1.4 meters when he is on the surface of the Earth. What is his initial velocity when he jumps?
Most of the time, however, Jack is in space. He is an asteroid miner: he looks for asteroids made out of useful metals. His job involves landing on asteroids and taking samples for analysis.
Jack knows that when he is on an asteroid, the local gravitational force can be very small. He must be careful -- if he stands up too fast or steps too hard, he could launch himself into space and never return.
One day, Joe lands on a small spherical asteroid of iron which has a radius of R=280 meters. What is the escape velocity from this body?
"Nope," says Joe, "it would be too dangerous to walk around on that asteroid. I might accidentally push myself away and never return."
What is the minimum radius a spherical asteroid of pure iron could have and still be "safe" for Jack? That is, how big must an iron asteroid be in order to prevent Jack from leaping off it by accident?
Homework Equations: F= -G(m1)(m2)/(radius)^2
G=(6.67*10^-11)
0^2 = initial velocity^2 + 2(-9.8m/s^2)(1.4m)
initial velocity =5.23m/s
Since ∫F= -G(m1)(m2)/(radius)^2 = G(m1)(m2)/radius
KE=(1/2)(m1)(velocity^2) = G(m1)(m2)/radius
m2=717.228*10^9 kg since it is a iron sphere
thus: ( (1/2) (velocity^2) = (6.67*10^-11) (717.228*10^9 kg) /250m
escape velocity=0.584m/s
For part 3, do I take 5.23m/s to substitute for velocity and solve for radius: (1/2)(m1)(velocity^2)=(m1)(m2)/radius
Most of the time, however, Jack is in space. He is an asteroid miner: he looks for asteroids made out of useful metals. His job involves landing on asteroids and taking samples for analysis.
Jack knows that when he is on an asteroid, the local gravitational force can be very small. He must be careful -- if he stands up too fast or steps too hard, he could launch himself into space and never return.
One day, Joe lands on a small spherical asteroid of iron which has a radius of R=280 meters. What is the escape velocity from this body?
"Nope," says Joe, "it would be too dangerous to walk around on that asteroid. I might accidentally push myself away and never return."
What is the minimum radius a spherical asteroid of pure iron could have and still be "safe" for Jack? That is, how big must an iron asteroid be in order to prevent Jack from leaping off it by accident?
Homework Equations: F= -G(m1)(m2)/(radius)^2
G=(6.67*10^-11)
0^2 = initial velocity^2 + 2(-9.8m/s^2)(1.4m)
initial velocity =5.23m/s
Since ∫F= -G(m1)(m2)/(radius)^2 = G(m1)(m2)/radius
KE=(1/2)(m1)(velocity^2) = G(m1)(m2)/radius
m2=717.228*10^9 kg since it is a iron sphere
thus: ( (1/2) (velocity^2) = (6.67*10^-11) (717.228*10^9 kg) /250m
escape velocity=0.584m/s
For part 3, do I take 5.23m/s to substitute for velocity and solve for radius: (1/2)(m1)(velocity^2)=(m1)(m2)/radius