A deep-space satellite is sent to orbit a distant planet with unknown mass. On arrival, the satellite begins its orbit and measures a gravitational pull from the planet of [tex]\(620 \, \text{N}\)[/tex]. If the satellite has a mass of [tex]\(450 \, \text{kg}\)[/tex] and orbits the planet at a radius of [tex]\(4.0 \times 10^6 \, \text{m}\)[/tex], what is the approximate mass of the planet? (Recall that [tex]\(G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2\)[/tex].)

A. [tex]\(3.3 \times 10^{23} \, \text{kg}\)[/tex]
B. [tex]\(2.9 \times 10^{22} \, \text{kg}\)[/tex]
C. [tex]\(7.8 \times 10^{23} \, \text{kg}\)[/tex]
D. [tex]\(4.5 \times 10^{24} \, \text{kg}\)[/tex]



Answer :

To determine the mass of the planet, we can use the formula for the gravitational force:

[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]

where:
- [tex]\( F \)[/tex] is the gravitational force (in Newtons, [tex]\( N \)[/tex]),
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the satellite,
- [tex]\( m_2 \)[/tex] is the mass of the planet,
- [tex]\( r \)[/tex] is the radius of the orbit (in meters).

In this problem:
- The gravitational pull, [tex]\( F = 620 \, N \)[/tex],
- The mass of the satellite, [tex]\( m_1 = 450 \, kg \)[/tex],
- The orbital radius, [tex]\( r = 4.0 \times 10^6 \, m \)[/tex],
- The gravitational constant, [tex]\( G = 6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \)[/tex].

We need to find the mass of the planet, [tex]\( m_2 \)[/tex].

First, rearrange the formula to solve for [tex]\( m_2 \)[/tex]:

[tex]\[ m_2 = \frac{F \cdot r^2}{G \cdot m_1} \][/tex]

Substitute the known values into the equation:

[tex]\[ m_2 = \frac{620 \, N \cdot (4.0 \times 10^6 \, m)^2}{6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \cdot 450 \, kg} \][/tex]

Calculate the value step by step:

1. Calculate [tex]\( r^2 \)[/tex]:
[tex]\[ (4.0 \times 10^6 \, m)^2 = 16 \times 10^{12} \, m^2 \][/tex]

2. Multiply [tex]\( F \)[/tex] by [tex]\( r^2 \)[/tex]:
[tex]\[ 620 \, N \cdot 16 \times 10^{12} \, m^2 = 9.92 \times 10^{15} \, N \cdot m^2 \][/tex]

3. Multiply [tex]\( G \cdot m_1 \)[/tex]:
[tex]\[ 6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \cdot 450 \, kg = 3.0015 \times 10^{-8} \, N \cdot m^2 / kg \][/tex]

4. Divide the result from step 2 by the result from step 3:
[tex]\[ m_2 = \frac{9.92 \times 10^{15}}{3.0015 \times 10^{-8}} \approx 3.305 \times 10^{23} \, kg \][/tex]

Thus, the approximate mass of the planet is:

[tex]\[ 3.305 \times 10^{23} \, kg \][/tex]

The answer is closest to option A:

[tex]\[ A. 3.3 \times 10^{23} \, kg \][/tex]