Question

(a) What is the magnitude of the force of gravity between Earth and Jupiter (take mass of Earth
ME = 6.00 1024 kg,
mass of Jupiter
MJ = 1.90 1027 kg,
and the distance between their centres
REJ = 5.89 108 m)?

b)At what point between Earth and Jupiter is the net force of gravity on a body by both Earth and Jupiter exactly zero?

Answer 1

Answer on Question 55489, Physics, Mechanics | Kinematics | Dynamics

Question:

a) What is the magnitude of the force of gravity between Earth and Jupiter (take mass of the Earth $M_{E} = 6.0 \cdot 10^{24} kg$, mass of Jupiter $M_{J} = 1.9 \cdot 10^{27} kg$, and the distance between their centres $R_{EJ} = 5.89 \cdot 10^{8} km$)?

b) At what point between Earth and Jupiter is the net force of gravity on a body by both Earth and Jupiter exactly zero?

Solution:

a) By the Newton's law of universal gravitation, the magnitude of the force of gravity between Earth and Jupiter is:

$$F_{EJ} = G \frac{M_{E} M_{J}}{R_{EJ}^{2}} = 6.67 \cdot 10^{-11} \frac{N \cdot m^{2}}{kg^{2}} \cdot \frac{6.0 \cdot 10^{24} kg \cdot 1.9 \cdot 10^{27} kg}{(5.89 \cdot 10^{11} m)^{2}} = 2.19 \cdot 10^{18} N.$$

b) Let $r$ be the distance from the center of the Jupiter to the point where the net force of gravity on a body is equal to zero and let $R_{EJ}$ is the distance between the centers of the Earth and Jupiter. Then the distance from the center of the Earth to the point of zero net force is $R_{EJ} - r$. By the Newton's law of universal gravitation, the force acting from the Jupiter on a body of mass $m$ at this point will be:

$$F = G \frac{M_{J} m}{r^{2}}.$$

The force acting from the Earth on a body of mass $m$ at this point will be (we use the relationship $M_{E} = \frac{1}{317} M_{J}$):

$$F = G \frac{\frac{1}{317} M_{J} m}{(R_{EJ} - r)^{2}}.$$

Thus, we equate this expressions:

$$\frac{1}{r^{2}} = \frac{\frac{1}{317}}{(R_{EJ} - r)^{2}},$$ $$\sqrt{317} (R_{EJ} - r) = r,$$ $$\sqrt{317} R_{EJ} = r \left(1 + \sqrt{317}\right),$$ $$r = \frac{\sqrt{317} R_{EJ}}{(1 + \sqrt{317})} = \frac{18}{19} R_{EJ} = \frac{18}{19} \cdot 5.89 \cdot 10^{11} \, \text{m} = 5.58 \cdot 10^{11} \, \text{m}.$$

Answer:

a) $F_{EJ} = 2.19 \cdot 10^{18} \, \text{N}$.

b) $r = 5.58 \cdot 10^{11} \, \text{m}$.

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