Is it correct the approach of fractions in finding the weight of an object in another planet?

Question

The problem is as follows:

A body has a weight of $w$ in the surface of the Earth. If the object is transported to a planet whose mass and radius is two times that of the Earth. Find its weight.

$\begin{array}{ll} 1.&4w\\ 2.&2w\\ 3.&\frac{w}{2}\\ 4.&\frac{w}{4}\\ 5.&w\\ \end{array}$

How should I calculate the weight of this object?.

On earth the only force acting in the object is given by the weight:

$F=mg=w$

And the gravitational force between two masses is given by:

$F=G\frac{m_1m_2}{r^2}$

Since it mentions that this object is moved to a planet which it has a radius which is two times that of the Earth and a mass double that of Earth then this becomes as:

$F_{2}=G\frac{m_1\cdot 2 m_2}{(2r)^2}=\frac{1}{2}G\frac{m_1m_2}{r^2}$

Therefore:

$w_{Planet}=\frac{1}{2}w_{Earth}$

But this doesn't make sense. What could I be doing wrong?. Shouldn't be the opposite. I mean two times that of the weight from Earth?. Can someone help me here?.

Answer 1

Wikipedia's Shell theorem article states:

Isaac Newton proved the shell theorem and stated that:

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.

Also, the density can vary, but only based on the distance from the center. Earth matches this ideal fairly closely, and I guess it's to be assumed so does this other planet. This means you can reasonably treat earth, and the other planet, to be point masses at their centers. Note this is why you're able to use the formula for the gravitational forces between $2$ masses, i.e., $F=G\frac{m_1m_2}{r^2}$, using the planets as a whole, instead of treating each bit of mass on its own.

Using this center point masses concept, the mass of the other planet being double means that, if the radius were the same, the weight on this planet would be double that of the same object on earth. However, the radius being double that of earth means, since the force of gravity is inversely proportional to the square of the distance, the gravity would be a factor of $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ that of earth. Thus, overall, any object's weight would be $2\left(\frac{1}{4}\right) = \frac{1}{2}$ that of earth, as you correctly determined, i.e., that $w_{\text{Planet}}=\frac{1}{2}w_{\text{Earth}}$.

Answer 2

If density were constant, (it isn't, but lets start there.)

Mass is proportional to volume. Volume increases with the cube of radius.

If we call this alien planet, planet $X,$ we can say about its mass: $m_X = 8m_E$

As for gravity. $g = G\frac {m_X w}{d^2}$

$d$ is $2\times$, $d^2$ is $4\times$, and $\frac{m_X}{d^2}$ is double earth gravity.

But density is not constant. $m_X = 2m_E.$ The density is $\frac 14$ earth density.

$g = G \frac{m_X w}{(2d)^2} = \frac 12 \frac{m_E w}{d^2}$

Gravity is less because the density is so much lower.

Answer 3

The new $g'$ would change by a factor of $M/R^2=2/2^2=1/2$, so $g'=g/2$ and $w'=w/2$.