Everything was great, but it just wasn’t a very exciting problem. The change in momentum of the spaceship was very small as it moved past the asteroid. Really, he could just have assumed a constant gravitational force and gotten pretty much the same answer. So, we came up with that astronaut scenario to spice things up a bit.
That introduced several challenging aspects: For one thing, since the asteroid is much smaller, it will also move. And how do you choose a good size for the time interval? Since both objects are at rest, they’ll be moving verrry slow at the outset. You can still calculate the changes with much smaller intervals, but that would take a lot of paper to solve. (Actually, you don’t have to use evenly spaced intervals, but let’s keep that constraint for fun.)
OK, Computer
First, let me show you how I’d solve this problem with a simple Python script. We’ll start with the standard model for the magnitude of the gravitational force between any two objects:
</p> <p><img alt="F equals G multipled by the product of m one and m two divided by r squared" class="responsive-image__image" src="https://i2.wp.com/www.ultimatepocket.com/wp-content/uploads/2019/10/could-an-astronaut-lost-in-space-use-gravity-to-get-around.jpg?w=640&ssl=1" srcset="https://i2.wp.com/www.ultimatepocket.com/wp-content/uploads/2019/10/could-an-astronaut-lost-in-space-use-gravity-to-get-around.jpg?w=640&ssl=1 1600w, https://i2.wp.com/www.ultimatepocket.com/wp-content/uploads/2019/10/could-an-astronaut-lost-in-space-use-gravity-to-get-around.jpg?w=640&ssl=1 1280w, https://i2.wp.com/www.ultimatepocket.com/wp-content/uploads/2019/10/could-an-astronaut-lost-in-space-use-gravity-to-get-around.jpg?w=640&ssl=1 1024w, https://i2.wp.com/www.ultimatepocket.com/wp-content/uploads/2019/10/could-an-astronaut-lost-in-space-use-gravity-to-get-around.jpg?w=640&ssl=1 768w, https://i2.wp.com/www.ultimatepocket.com/wp-content/uploads/2019/10/could-an-astronaut-lost-in-space-use-gravity-to-get-around.jpg?w=640&ssl=1 640w" sizes="100vw" data-recalc-dims="1"></p> <p>
Illustration: Rhett Allain
In this expression, m1 and m2 are the masses of the two objects—let’s say the astronaut is 100 kg and the rock is 200 kg. Also, r is the distance between them (3 meters), and G is the gravitational constant, which has a value of 6.67 x 10–11 N×m2/kg2. First off, notice that G is a tiny number: 0.0000000000667. Which means the gravitational force between two objects is very small unless one of them is MASSIVE. So that’s not promising …
But let’s put the numerical calculation into a Python script. (Here’s the code.) I’m going to pick a time interval of 100 seconds and let stuff update until the center of the astronaut and rock are 0.5 meters apart—as a rough approximation of when they’ll make surface contact. I won’t show an animation of this, because it’s about as interesting as watching grass grow. But below is a plot of the position of both objects until they meet.
Cool, huh? From this, you can see that the two objects move toward each other, but the astronaut travels farther than the asteroid, and it takes 11.03 hours to make contact. (The plot above shows the time in seconds: “4e+4” means 4 × 10^4, or 40,000 seconds, which is about 11.1 hours.) That’s not too crazy.
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