calculating you weight if you $100\text{Km}$ above the earth etc.

Question

Qualitatively, how would your weight be different if you weighed yourself:

1. $100\text{Km}$ above the earth?
2. $100\text{Km}$ above the moon?
3. near the equator if the earth's rotation were $10\text{h}$?

I don't know how to even start this question.

Answer 1

OK, you know the gravitational law:

$$F = \frac{G M m}{r^2} $$

where $M$ is the mass of the earth and $m$ is your mass.

You weigh, on the surface of the earth

$$F_0 = \frac{G M m}{R_e^2} $$

where $R_e$ is the radius of the earth. Your weight at a height $h$ above is

$$F_h = \frac{G M m}{(R_e+h)^2} $$

Can you now compare these weights?

Answer 2

1. We know that the value of Earth's gravitational acceleration on the surface is given as $$g=\frac{Gm_e}{R_e^2}=9.81$$

Where, $M_e=\text{mass of Earth}=6\times 10^{24}\ kg$ & $R_e=\text{radius of Earth}=6400\ km$

Consider an object of wight $W$ on the Earth's surface

Then, the weight of the object at a height $h$ from the Earth's surface is given by Newton's Law of Gravitation $$mg'=\frac{GM_em}{(R_e+h)^2}$$ $$=\frac{GM_em}{R_e^2\left(1+\frac{h}{R_e}\right)^2}$$ $$=\frac{mg}{\left(1+\frac{h}{R_e}\right)^2}$$

Hence, if $mg=\text{weight on the Earth's surface}=W$

then the weight $W'$ of the object at the height $h=100\ km$ is $$W=\frac{W}{\left(1+\frac{100}{6400}\right)^2}$$ then the weight $W'$ of the object at the height $h=100\ km$ is $$W'=\frac{W}{\left(1+\frac{100}{6400}\right)^2}$$ $$W'=\frac{W}{\left(1+\frac{100}{6400}\right)^2}$$$$=\left(\frac{64}{65}\right)^2W$$ $$\color{blue}{W'=\frac{4096 W}{4225}=\frac{4096 }{4225}\times \text{(weight on the surface of Earth)}}$$

2. Now, for moon, take the radius of moon $R_m=1737\ km$ & acceleration due to Moon's gravity as $\frac{g}{6}$ where, $g=\text{Earth's gravitational acceleration}$

Hence, at the height $h=100\ km$ from the surface of moon the weight $W''$ of object

$$W''=\frac{m\frac{g}{6}}{\left(1+\frac{100}{1737}\right)^2}$$ $$W''=\frac{\frac{W}{6}}{\left(1+\frac{100}{1737}\right)^2}$$$$=\frac{W}{6}\left(\frac{1737}{1837}\right)^2$$ $$\color{blue}{W''=\frac{3017169 W}{20247414}\approx (0.149)\times \text{(weight on the surface of Earth)}}$$

3. We know that gravitational acceleration of Earth $g'$ due to Earth's rotation velocity $\omega$ is given as $$\color{red}{g'=g-2\omega^2R_e\cos^2\lambda}$$ Where, $\lambda=\text{angle of latitude of the point from equator}$ & $\omega=\text{angular velocity (in rad/sec)}$

But, at the equator plane $\lambda=0$ hence, the weight of the object $$W'=m(g-2\omega^2R_e\cos^20)$$ $$W'=m(g-2\omega^2R_e)$$

& for rotation time $10$ hrs we get $$\omega=\frac{2\pi}{10\times 3600}=\frac{\pi}{18000}\ rad/sec$$

Now, setting the values of mass, $m$, $g=9.81$ at the pole, $\omega=\frac{\pi}{18000}\ rad/sec$ & $R_e=6400000\ m$

weight can be calculated as $$\color{blue}{W'=m(g-2\omega^2R_e)}$$ I hope you may solve the remaining part.

Answer 3

I think the key to answering this question is the word qualitatively.

The question is not asking you to calculate how much you will weigh at these various places. It is asking you whether you would weigh the same, less, or more than you do now.