Escape velocity of an iron asteroid

In summary, Jack can jump upwards a distance of 1.4 meters when he is on the surface of the Earth, with an initial velocity of 5.23m/s. However, most of the time he is in space as an asteroid miner. When on an asteroid, he must be careful not to launch himself into space due to the low local gravitational force. For a small spherical iron asteroid with a radius of 280 meters, the escape velocity would be 0.584m/s, making it too dangerous for Jack to walk around on. The minimum radius for a spherical asteroid of pure iron to be "safe" for Jack would be 78.8 * 10^3 meters. This can be calculated using the formula
  • #1
ac7597
126
6
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
 
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  • #2
ac7597 said:

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
Yes. Of course the mass depends on radius as well.
 
  • #3
since (1/2)(m1)(velocity^2)=G(m1)(m2)/radius
m2=(4/3)(pi)(radius^3)(7.874 g/cm^3)
thus: (1/2)(5.23m/s)^2=(6.67*10^-11)( (4/3)(pi)(radius^3)(7.874 g/cm^3) ) /(radius)
radius= 32.2 *10^3
Is this correct?
 
  • #4
correction: radius=78.8 * 10^3
Is the units correct in terms of meter?
 
  • #5
Can you think of a way to check that your result is correct? Earlier you calculated the escape velocity for a small sphere of known radius, so perhaps you can do the same for this new sphere radius and compare that value to something?

By the way, for the first part of the problem you write 250m in your work, but the description mentions 280m.
 
  • #6
ac7597 said:
since (1/2)(m1)(velocity^2)=G(m1)(m2)/radius
m2=(4/3)(pi)(radius^3)(7.874 g/cm^3)
thus: (1/2)(5.23m/s)^2=(6.67*10^-11)( (4/3)(pi)(radius^3)(7.874 g/cm^3) ) /(radius)
radius= 32.2 *10^3
Is this correct?
No. The calculation should give the right result if you express the density of iron in kg/m^3. You must have mad e some other error as well. The larger number you give in post #4 is incorrect as well.
The calculation is easier if you compute a formula of how the escape speed depends on the radius first.
 

FAQ: Escape velocity of an iron asteroid

1. What is the escape velocity of an iron asteroid?

The escape velocity of an iron asteroid is the minimum speed required for an object to escape the gravitational pull of the asteroid and not fall back to its surface. It is calculated using the mass and radius of the asteroid, and is typically measured in kilometers per second.

2. How is the escape velocity of an iron asteroid different from other types of asteroids?

The escape velocity of an iron asteroid is usually higher than other types of asteroids due to its higher density and mass. This means that it takes more energy to escape the gravitational pull of an iron asteroid compared to a less dense asteroid of the same size.

3. What factors affect the escape velocity of an iron asteroid?

The escape velocity of an iron asteroid is primarily affected by its mass and radius, as well as the gravitational constant of the asteroid. Other factors that may play a role include the asteroid's rotation speed and the presence of other nearby objects with significant gravitational pull.

4. Can the escape velocity of an iron asteroid change?

Yes, the escape velocity of an iron asteroid can change over time. This can occur if the asteroid's mass or radius changes due to collisions with other objects, or if its rotation speed changes. It can also be affected by changes in the asteroid's environment, such as the presence of other gravitational forces.

5. Why is the escape velocity of an iron asteroid important to understand?

The escape velocity of an iron asteroid is important to understand because it can impact the trajectory and behavior of the asteroid. It also plays a role in determining the potential for the asteroid to collide with other objects or be affected by other gravitational forces. Additionally, knowing the escape velocity can provide valuable information for potential asteroid mining or spacecraft missions to the asteroid.

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