The escape velocity of a bullet from the surface of planet Y is 1983.0 m/s. Calculate the escape velocity from the surface of the planet X if the mass of planet X is 1.31 times that of Y, and its radius is 0.903 times the radius of Y.
Escape velocity is given by:\[v _{e}=\sqrt{\frac{ 2GM }{ r }}\]where ve is the escape velocity; G is the gravitational constant; M is the mass of the planet; and r is the radius of the planet.
Do you see a way of using the formula to solve your problem?
yes, but the two planets kind of confuse me
Why do they confuse you?
Like how "X is 1.31 times that of Y, and its radius is 0.903 times the radius of Y"
You don't need to know the actual mass of Planet Y or X; and you don't need to know the actual radius of either planet.
So you know that the mass of planet Y=M. You also note that the mass of Planet X =1.31M. You also know that radius of planet X = 0.903r
Now we can set up a ratio:\[\frac{ v _{X} }{ v _{Y} }=\frac{ \sqrt{\frac{ 2GM _{X} }{ r _{X} }} }{ \sqrt{\frac{ 2GM _{Y} }{ r _{Y}}} }=\sqrt{\frac{ r _{Y}M _{X} }{ r _{X}M _{Y} }}\]
Let's say that Mx=aMy and rx=bry. If we put that in my equation we get:\[\frac{ v _{X} }{ v _{Y} }=\sqrt{\frac{ r _{Y}aM _{Y} }{ br _{Y} M _{Y}}}=\sqrt{\frac{ a }{ b }}\]
That means that vX is given by the following:\[v _{X}=v _{Y}\sqrt{\frac{ a }{ b }}\]
Do you see how to solve the problem now?
Let me try and see this one second...
vx=1983.0sqrt(...) I am confused as to what a and b would be
it wouldn't be 1.31 and .903 would it?
You see where I typed that Mx=aMy? That's the same as saying the mass of Planet X is 1.31 times the mass of Planet Y. In other words, Mx=1.31My.
Yes, those would be a and b.
I'm out of here, but if you get an answer that looks like 2388.4m/s, you've done things correctly.
Okay, thank you for your help!
I got it! Thank you!
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