How do I find the weight of an object in a different planet?

Jun 2017
399
6
Lima, Peru
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?.
 

skeeter

Math Team
Jul 2011
3,353
1,842
Texas
It makes perfect sense. Weight of a mass on the surface is proportional to the mass of the planet and inversely proportional to the square of that planet's radius

On Earth, $w = k \cdot \dfrac{M_e}{R_e^2}$, where $k$ is the constant of proportionality.

On the planet with twice the mass & radius of the Earth ...

$w = k \dfrac{2M_e}{(2R_e)^2} = k \dfrac{2M_e}{4R_e^2} = \dfrac{1}{2} \cdot k \dfrac{M_e}{R_e^2} = \dfrac{1}{2}w$

... the inverse proportionality of the radius squared is the driving factor
 
Jun 2017
399
6
Lima, Peru
It makes perfect sense. Weight of a mass on the surface is proportional to the mass of the planet and inversely proportional to the square of that planet's radius

On Earth, $w = k \cdot \dfrac{M_e}{R_e^2}$, where $k$ is the constant of proportionality.

On the planet with twice the mass & radius of the Earth ...

$w = k \dfrac{2M_e}{(2R_e)^2} = k \dfrac{2M_e}{4R_e^2} = \dfrac{1}{2} \cdot k \dfrac{M_e}{R_e^2} = \dfrac{1}{2}w$

... the inverse proportionality of the radius squared is the driving factor
Okay. But when you put an object on let's say Jupiter, wouldn't its weight be more than what is it on Earth?.

Let's say a $46\,g$ golf ball. On the surface of the Earth, its weight would be:

$46\cdot10^-3\cdot 9.8 = 0.4508\,N$

On Jupiter (assuming its mass is $1.898 \times 10^{27}\,kg$ and a radius of $69911\,km$)

$F=(6.67408 \cdot 10^{-11})\left(\frac{46\cdot10^-3\cdot 1.898 \times 10^{27}}{(69911\cdot 10^3)^2} \right)$

$F\approx 1.1922\,N$

The weight increases. Why is it increasing?. It's almost 1.6 times that of the weight on the surface of Earth. Or could it be that am I missunderstanding something?.
 

skeeter

Math Team
Jul 2011
3,353
1,842
Texas
First of all, $\dfrac{1.1922}{0.4508} = 2.64$, so the golf ball is about 2.6 times heavier in Jupiter than on Earth.

The value of the gravitational field on the surface of any planet X is $g_X = \dfrac{GM_X}{(R_X)^2}$

On Earth, $g_e = \dfrac{GM_e}{(R_e)^2}$, which yields the familiar $9.8 \, m/s^2$

comparing the two ...

$\dfrac{g_X}{g_e} = \dfrac{M_X (R_e)^2}{M_e (R_X)^2} = \dfrac{M_X}{M_e} \cdot \left(\dfrac{R_e}{R_X} \right)^2$

Let planet X be Jupiter. Looking at each fraction individually ...

$\dfrac{M_J}{M_e} \approx 318$

$\dfrac{R_e}{R_J} \approx 0.09$

$318 \cdot 0.09^2 \approx 2.6$, so gravitational acceleration on Jupiter's surface is 2.6 times greater than Earth's. Nothing mysterious here, it's just the way the numbers work out. In this case, Jupiter's much greater mass is the reason for the greater weight.
 
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