created at April 22, 2021

Jump on the Moon

It's interesting fact that Moon has always the same orientation relative to the Earth. However, let's consider following situation:

If we are staying on the visible Moon side, then Earth attraction allow us to jump higher in comparison with the other side (back side). On the back side Earth attraction will prevent us from jumping higher.


How much higher will be the jump on the visible Moon side?

Let's designate forces acting on the astronaut on the Moon:

F1 - Moon attraction, F2 - Earth attraction

According to the Newton's law of universal gravitation, two bodies with masses m1 and m2 and distance R between them attract each other using force:

F=Gm1m2R2[N]F=G \frac{m1\cdot m2}{R^{2}}\quad [N]

G - gravitational constant (G = 6.67408*10-11 m3/(kg*s2)), (R - distance between center of gravities of the bodies). Attraction forces from Moon and Earth sides are:

F1=GM1md12;F2=GM2md22;F1=G \frac{M1\cdot m}{d1^{2}};\quad F2=G \frac{M2\cdot m}{d2^{2}};

where M1 and M2 - Moon and Earth masses accordingly, d1 and d2 - distances between the astronaut and Moon's and Earth's center of gravities accordingly, m - astronaut's mass.

Considering force F to be negative (positive direction is from the center of the planet and planet's attraction acts towards the center), the jump height is:

H=12mV02F[m]H=\frac{1}{2}\cdot \frac{m\cdot V_{0}^{2}}{F}\quad [m]

where V0 - some initial jump vertical velocity. The resulting force on the visible (frontal) side:


Resulting force on the back side:


Substituting Ff and Fb into the (1), we get jumping heights on the front and back sides:

Hf=12V02G(M1d12M2d22)H_{f}=\frac{1}{2}\cdot \frac{V_{0}^{2}}{G\cdot (\frac{M1}{d1^{2}} - \frac{M2}{d2^{2}})}
Hb=12V02G(M1d12+M2d22)H_{b}=\frac{1}{2}\cdot \frac{V_{0}^{2}}{G\cdot (\frac{M1}{d1^{2}} + \frac{M2}{d2^{2}})}

then the ratio of both heights is:

r=(M1d12+M2d22)(M1d12M2d22)r=\frac{(\frac{M1}{d1^{2}} + \frac{M2}{d2^{2}})}{(\frac{M1}{d1^{2}} - \frac{M2}{d2^{2}})}

Substituting Moon's and Earth's masses (M1 = 7.3477Е22 kg, M2 = 5.9726E24 kg), and distances d1 ≈ 1737 km, d2 ≈ 384000 km, we get:

r=(7.3477E2217372+5.9726E243840002)(7.3477E22173725.9726E243840002)=1.0033r=\frac{(\frac{7.3477E22}{1737^{2}} + \frac{5.9726E24}{384000^{2}})}{(\frac{7.3477E22}{1737^{2}} - \frac{5.9726E24}{384000^{2}})}=1.0033

So, jump on the frontal side of the Moon will be only 0.33 percent higher than on the back side.