Question

1 multiple choice 1 point calculate the gravitational force between the sun and the earth using the following values: g (gravitational constant) = 6.674 x 10⁻¹¹ newton-meter²/kg² mass of the sun = 1.989 x 10³⁰ kg mass of the earth = 5.972 x 10²⁴ kg distance = 1.495 x 10¹¹ meters 5.31 x 10³² newtons 3.55 x 10²² newtons 7.94 x 10⁴³ newtons 5.30 x 10³⁰ newtons

Explanation:

Step1: Recall the gravitational force formula

The formula for gravitational force between two objects is \( F = G\frac{m_1m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers.

Step2: Substitute the given values into the formula

Given \( G = 6.674\times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2 \), \( m_1 = 1.989\times 10^{30}\ \text{kg} \) (mass of Sun), \( m_2 = 5.972\times 10^{24}\ \text{kg} \) (mass of Earth), and \( r = 1.495\times 10^{11}\ \text{m} \).

First, calculate the numerator: \( Gm_1m_2 = 6.674\times 10^{-11} \times 1.989\times 10^{30} \times 5.972\times 10^{24} \)

Multiply the coefficients: \( 6.674\times 1.989\times 5.972 \approx 6.674\times 11.86 \approx 79.1 \)

Multiply the exponents: \( 10^{-11} \times 10^{30} \times 10^{24} = 10^{-11 + 30 + 24} = 10^{43} \)

So the numerator is approximately \( 79.1\times 10^{43} = 7.91\times 10^{44} \)

Now calculate the denominator: \( r^2 = (1.495\times 10^{11})^2 = (1.495)^2\times 10^{22} \approx 2.235\times 10^{22} \)

Step3: Divide the numerator by the denominator

\( F = \frac{7.91\times 10^{44}}{2.235\times 10^{22}} \)

Divide the coefficients: \( \frac{7.91}{2.235} \approx 3.54 \)

Divide the exponents: \( 10^{44 - 22} = 10^{22} \)

So \( F \approx 3.54\times 10^{22}\ \text{N} \), which is approximately \( 3.55\times 10^{22}\ \text{N} \)

Answer:

B. \( 3.55\times 10^{22} \) newtons