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anonymous
 3 years ago
23 )If a planet has a radius 10 times the radius of the earth and five times the mass, what is the value of g on this planet?
A) 9.8 m/sec^2
B) 6.67x10^11 Nxm^2/kg^2
C) 1/20 of G on earth
D) 20 times the value of g on earth
E) none of these
i said 1 /20 of g on earth is that correct?
and also another question
.What would your weight be at the center of the planet described in the first question? Greater than at the surface of the planet .Slightly less than at the surface of the planet the same as at the surface or 0 N?
i said 0 n
anonymous
 3 years ago
23 )If a planet has a radius 10 times the radius of the earth and five times the mass, what is the value of g on this planet? A) 9.8 m/sec^2 B) 6.67x10^11 Nxm^2/kg^2 C) 1/20 of G on earth D) 20 times the value of g on earth E) none of these i said 1 /20 of g on earth is that correct? and also another question .What would your weight be at the center of the planet described in the first question? Greater than at the surface of the planet .Slightly less than at the surface of the planet the same as at the surface or 0 N? i said 0 n

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JFraser
 3 years ago
Best ResponseYou've already chosen the best response.0plug into the equation for universal gravitation

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3\[\LARGE g_{earth}=\frac{GM_e}{R_e^2}....(Equation~1)\] \[\LARGE g_{planet}=\frac{G \times 5M_e}{10R_e^2}\] 5 and 10 will cancel out,you will thus obtain a relation between \[\large g_{earth} ~and ~g_{planet}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.3Yes you are right with 1/20 times g on earth.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0comparisions become easier when you think of proportions. \[g={GM\over R^2}\implies g\propto{M\over R^2}\\ {g_1\over g_2}={M_1\over M_2}\times\left(R_2\over R_1\right)^2 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Seems the first has been answered, and as for the second question, yes, 0N is correct since at the center you would have equal forces of gravity on all sides.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks alot everybody
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