Question

problem 7:
enceladus is a moon of saturn. this moon is one of the main candidates to have life other than earth in the solar system. let us find the tidal force on enceladus by saturn. the first thing is to take a 1 kg mass and find the force on this mass at the point of enceladus nearest saturn. therefore, one needs the mass of saturn, ms = 5.68 × 10²⁶ kg, the distance from the center of saturn to the center of enceladus, r = 238,000 km, and the radius of enceladus, r = 252,100 m. the next thing is to take this same 1 kg mass, place it on enceladus as far as possible from saturn and find the force on it by saturn. the difference of these two forces is the tidal force. find the tidal force

Explanation:

Step1: Calculate the force at the nearest point

Using Newton's law of gravitation \(F = G\frac{Mm}{d^{2}}\), where \(G = 6.67\times10^{- 11}\space Nm^{2}/kg^{2}\), \(M = 5.68\times10^{26}\space kg\), \(m = 1\space kg\), and \(d=R - r\). First, convert \(R = 238000\space km=238000\times10^{3}\space m\). Then \(d=(238000\times10^{3}-252100)\space m\approx2.3775\times10^{8}\space m\).

$$F_{1}=G\frac{Mm}{(R - r)^{2}}=6.67\times 10^{-11}\times\frac{5.68\times 10^{26}\times1}{(2.3775\times 10^{8})^{2}}$$

$$F_{1}=6.67\times 10^{-11}\times\frac{5.68\times 10^{26}}{5.6526\times 10^{16}}$$

$$F_{1}\approx6.67\times10^{-11}\times1.005\times10^{10}$$

$$F_{1}\approx6.7\space N$$

Step2: Calculate the force at the farthest point

Using \(d = R + r\), \(d=(238000\times10^{3}+252100)\space m\approx2.3825\times10^{8}\space m\)

$$F_{2}=G\frac{Mm}{(R + r)^{2}}=6.67\times 10^{-11}\times\frac{5.68\times 10^{26}\times1}{(2.3825\times 10^{8})^{2}}$$

$$F_{2}=6.67\times 10^{-11}\times\frac{5.68\times 10^{26}}{5.676\times 10^{16}}$$

$$F_{2}\approx6.67\times10^{-11}\times1.001\times10^{10}$$

$$F_{2}\approx6.68\space N$$

Step3: Calculate the tidal force

The tidal force \(F_{tidal}=F_{1}-F_{2}\)

$$F_{tidal}\approx6.7 - 6.68$$

$$F_{tidal}\approx0.02\space N$$

Answer:

The tidal force is approximately \(0.02\space N\)