The acceleration due to gravity at the north pole of Neptune is approximately 11.2m/s2. Neptune has mass 1.02×1026kg and radius 2.46×104km and rotates once around its axis in about 16 h. (a) What is the gravitational force on a 3.00-kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune’s equator? (Note that Neptune’s "surface" is gaseous, not solid, so it is impossible to stand on it.)

[tex]F=\frac{Gmm}{R^{2} }\\ G=6.67*10^{-11}\frac{Nm^{2} }{kg^{2} }\\ F=\frac{6.67*10^{-11}*1.02*10^{26}*3.00 }{(2.46*10^{4} *1000)^{2} }\\ F=33.7N[/tex]

For (b) part Weight

[tex]W=F-\frac{mv^{2} }{R}\\ as\\T=\frac{2*\pi *r}{V}\\ V=\frac{2*\pi *r}{T}\\ V=\frac{2*(3.14)*(2.46*10^{4} *1000)}{16*3600} \\V=2683 \frac{m}{s}[/tex]

V=2683 m/s

[tex]W=33.7-\frac{3*(2683)^{2} }{2.46*10^{4}*1000 }\\ W=32.8N[/tex]

annyksl annyksl · Answer 2 · 2022-03-29T15:02:13+02:00

The gravitational force on the object at Neptune's north pole is equal to 33.7N. The weight of this object is equal to 32.8N.

How can we arrive at this result?

First, we must find the gravitational force.

This will be done with the equation [tex]F=\frac{Gmm}{R^2}[/tex]

For "G" we will adopt the value of [tex]6.67*10^-^1^1\frac{Nm^2}{Kg^2}[/tex]

After that, we can use the equation as follows:

[tex]F=\frac{6.67*10^-^1^1*1.02*10^2^6*3}{(2.46*10^4*1000)^2} \\F=33.7N[/tex]

With the value of the gravitational force it is possible to find the apparent weight of the object through the equation: [tex]W= F-\frac{mv^2}{R}[/tex]

However, to use this equation, we must first find the value of velocity. This will be done with the equation: [tex]v= \frac{2*\pi*r }{t}[/tex]

Solving the velocity equation we get:

[tex]v= \frac{2*\pi*2.46*10^4*1000}{16*3600} \\v= 2683 \frac{m}{s}[/tex]

With this value, we can calculate the apparent weight equation.

[tex]W=33.7-\frac{3*2683^2}{2.46*10^4*1000} = 32.8N[/tex]

More information about weight is in the link:

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