The electronic scale measures based on the normal force the scale provides to the object. This in turn is based on the force of gravity on the object by the earth.

$\displaystyle F_G=-G\frac{m_1 m_2}{r^2}$

As height above sea level increases, as does $\displaystyle r$, the distance to the center of the Earth. As $\displaystyle r$ increases, $\displaystyle F_G$ decreases. This would decrease the normal force which would decrease the reading on the scale.

First, estimate the acceleration due to gravity close to the moon's surface:

$\displaystyle F=ma$

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

Combining equations and solving for the acceleration:

$\displaystyle a_{mass}=-G\frac{m_{moon}}{r_{moon}^2}$

Converting $\displaystyle km$ to $\displaystyle m$ and pugging in values:

$\displaystyle a_{mass}=-6.67*10^{-11}\frac{7.3*10^{22}}{(1737*10^3)^2}$

$\displaystyle a_{mass}=-1.61\frac{m}{s^2}$

Using

$\displaystyle F_{total}=F_1+F_2+...$

$\displaystyle F_{gravity}=ma_{gravity}$

$\displaystyle F_{total}=kx+ma_{gravity}$

Solving for $\displaystyle k$

$\displaystyle k=\frac{F_{total}-ma_{gravity}}{x}$

Converting $\displaystyle cm$ to $\displaystyle m$ and plugging in values:

$\displaystyle k=\frac{0-15*-1.61}{.01}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$\displaystyle k=\frac{2415N}{m}$

Using

$\displaystyle F=ma$

$\displaystyle F_{total}=F_1+F_2+...$

Combining equations:

$\displaystyle ma=F_{tension}+mg$

Solving for $\displaystyle F_{tension}$ and plugging in values:

$\displaystyle F_{tension}=5*2-5*(-9.8)$

$\displaystyle F_{tension}=59N$

Finding total distance from center of Earth to astronaut:

$\displaystyle 200km+6371km=6571km$

Converting to meters

$\displaystyle 6.571*10^6m$

Using Universal Gravitation equation:

$\displaystyle F_G=-G\frac{m_1m_2}{r^2}$

Plugging in values:

$\displaystyle F_G=-6.67*10^{-11}\frac{5.972*10^{24}*110}{(6.571*10^6)^2}$

$\displaystyle |F_G|=1015N$

Using

$\displaystyle F=ma$

and

$\displaystyle |F_G|=G\frac{m_1 m_2}{r^2}$

Combining equations

$\displaystyle m_1a=G\frac{m_1m_2}{r^2}$

Solving for $\displaystyle a$

$\displaystyle a=G\frac{m_2}{r^2}$

Plugging in values:

$\displaystyle a=6.67*10^{-11}\frac{1.89*10^{27}}{(1670*10^6)^2}$

$\displaystyle a=4.52*10^{-2}\frac{m}{s^2}$

First, estimate the acceleration due to gravity close to Pluto's surface:

$\displaystyle F=ma$

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

Combining equations and solving for the acceleration:

$\displaystyle a_{mass}=-G\frac{m_{pluto}}{r_{pluto}^2}$

Plugging in values:

$\displaystyle a_{mass}=-6.67*10^{-11}\frac{1.31*10^{22}}{(1.19*10^6)^2}$

$\displaystyle a_{mass}=-.617\frac{m}{s^2}$

Using

$\displaystyle F_{total}=F_1+F_2+...$

$\displaystyle F_{gravity}=ma_{gravity}$

$\displaystyle F_{total}=kx+ma_{gravity}$

Solving for $\displaystyle k$

$\displaystyle k=\frac{F_{total}-ma_{gravity}}{x}$

Converting $\displaystyle cm$ to $\displaystyle m$ and plugging in values:

$\displaystyle k=\frac{0-27*(-.617)}{.04}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$\displaystyle k=\frac{416N}{m}$

First, estimate the acceleration due to gravity close to the martian surface:

$\displaystyle F=ma$

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

Combining equations and solving for the acceleration:

$\displaystyle a_{mass}=-G\frac{m_{mars}}{r_{mars}^2}$

Plugging in values:

$\displaystyle a_{mass}=-6.67*10^{-11}\frac{639*10^{21}}{(3.39*10^6)^2}$

$\displaystyle a_{mass}=-3.71\frac{m}{s^2}$

Using

$\displaystyle F_{total}=F_1+F_2+...$

$\displaystyle F_{gravity}=ma_{gravity}$

$\displaystyle F_{total}=kx+ma_{gravity}$

Solving for $\displaystyle k$

$\displaystyle k=\frac{F_{total}-ma_{gravity}}{x}$

Converting $\displaystyle cm$ to $\displaystyle m$ and plugging in values:

$\displaystyle k=\frac{0-65*(-3.71)}{.055}$

(Since the mass is at rest, the acceleration and thus the net force is zero)

$\displaystyle k=\frac{4385N}{m}$

Setting universal gravitation equal to the surface gravitational approximation

$\displaystyle -G\frac{m_1m_2}{r^2}=-m_2*g_a$

Where $\displaystyle g_a$ will be the surface gravitational acceleration constant

Solving for $\displaystyle g_a$

$\displaystyle g_a=G\frac{m_1}{r^2}$

Plugging in values

$\displaystyle g_a=6.67*10^{-11}\frac{6.3*10^{20}}{250,000^2}$

$\displaystyle g_a=.672\frac{m}{s}$

Setting universal gravitation equal to the gravitational approximation

$\displaystyle -G\frac{m_1m_2}{r^2}=-m_2*g_a$

Where $\displaystyle g_a$ will be the surface gravitational acceleration constant

Solving for $\displaystyle g_a$

$\displaystyle g_a=G\frac{m_1}{r^2}$

Converting $\displaystyle km$ to $\displaystyle m$ and plugging in values

$\displaystyle g_a=6.67*10^{-11}\frac{1.1*10^{20}}{200,000^2}$

$\displaystyle g_a=.183\frac{m}{s^s}$

The gravity equation is:

$\displaystyle g=\frac{GM}{R^2}$, where $\displaystyle G$ is the universal gravitation constant, $\displaystyle M$ is the mass of the planet, and $\displaystyle R$  is the radius of planet. To find a ratio between the gravity on the planet Zina and our planet, we need to plug in the relevant information:

$\displaystyle \frac{g_{Zina}}{g_{Earth}}=\frac{\frac{GM_{Zina}}{R^2_{Zina}}}{\frac{GM_{Earth}}{R^2_{Earth}}}=\frac{\frac{G*2*M_{Earth}}{3^2*R^2_{Earth}}}{\frac{GM_{Earth}}{R^2_{Earth}}}$

$\displaystyle =\frac{\frac{G*2}{3^2}}{\frac{G*1}{1}}=\frac{2}{9}.$

Remember, we are doing a ratio, so if we reference everything in terms of Earth numbers, the bottom fraction becomes much simpler and cancels out with the top fraction. Also, the gravitational constant is constant, so it cancels out. Therefore, the only important information are the ratios the mass and radius differ from Earth's.