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anonymous
 one year ago
An astronaut has weight W on Earth. The astronaut travels to a planet that has a mass 8 times greater than Earth's mass, and its radius is 2 times greater than Earth's radius. What is the astronaut's weight on the planet?
anonymous
 one year ago
An astronaut has weight W on Earth. The astronaut travels to a planet that has a mass 8 times greater than Earth's mass, and its radius is 2 times greater than Earth's radius. What is the astronaut's weight on the planet?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1from Newton's Law for gravitation, we know that: \(\huge F = \frac{G \ M \ m}{r^2}\) [F is the proxy for weight] so: \(\huge G = \frac{F \ r^2 }{\ Mm} = const.\) so \(\huge \frac{F_1 \ r_1^2 }{\ M_1m} = \frac{F_2 \ r_2^2 }{\ M_2m}\) thus: \(\huge \frac{F_1 \ r_1^2 }{\ M_1} = \frac{F_2 \ r_2^2 }{\ M_2}\) that's weird, m is irrelevant...... \(\huge F_2 = F_1 (\frac{M_2}{M_1})( \frac{r_1}{r_2})^2\) so his actual mass is meaningless.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1"duh" we are comparing things so it ought to cancel out...

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[W_\oplus=F_\oplus = G\frac{M_\oplus m}{R_\oplus^2}=mg_\oplus\]\[W_x=F_x = G\frac{M_x m}{R_x^2}=mg_x\] \[R_x=2R_\oplus,\qquad M_x=8M_\oplus \] \[W_x=F_x = G\frac{(8M_\oplus) m}{(2R_\oplus)^2}=\dots%\left(\frac84\right)G\frac{M_\oplus m}{R_\oplus^2}=2W_\oplus \]
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