Question

jupiter has a mass of 1.90 × 10²⁷ kg. one of its moons, europa, has a mass of 4.80 × 10²² kg. the average distance from jupiter’s center to europa’s center is 6.71 × 10⁸ m. calculate the magnitude of the gravitational force jupiter exerts on europa when europa is at its average distance from jupiter. choose 1 answer: a) 1.35 × 10²² n b) 2.03 × 10³² n c) 9.07 × 10⁶⁰ n

Explanation:

Step1: Recall Gravitational Force Formula

The formula for gravitational force is \( F = G\frac{m_1m_2}{r^2} \), where \( G = 6.674\times 10^{-11}\, \text{N}\cdot\text{m}^2/\text{kg}^2 \), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers.

Step2: Identify Given Values

\( m_1 = 1.90\times 10^{27}\, \text{kg} \) (mass of Jupiter), \( m_2 = 4.80\times 10^{22}\, \text{kg} \) (mass of Europa), \( r = 6.71\times 10^{8}\, \text{m} \).

Step3: Substitute Values into Formula

First, calculate the numerator: \( Gm_1m_2 = 6.674\times 10^{-11} \times 1.90\times 10^{27} \times 4.80\times 10^{22} \)

$$
LATEXBLOCK0
$$

Then, calculate the denominator: \( r^2=(6.71\times 10^{8})^2 = 6.71^2\times 10^{16}\approx 45.02\times 10^{16}=4.502\times 10^{17} \)

Step4: Divide Numerator by Denominator

\( F=\frac{6.097\times 10^{39}}{4.502\times 10^{17}} \approx \frac{6.097}{4.502}\times 10^{39 - 17}\approx 1.35\times 10^{22}\, \text{N} \)

Answer:

A. \( 1.35 \times 10^{22}\, \text{N} \)