Let's break down each part of this problem systematically using Newton's law of universal gravitation and the formula for acceleration.

Given:

• Mass of the satellite: $m = 461 \, \text{kg}$
• Distance from the center of the Earth: $r = 2.19 \, R_E$, where $R_E$ is the radius of the Earth $R_E = 6.37 \times 10^6 \, \text{m}$
• Mass of the Earth: $M_E = 5.972 \times 10^{24} \, \text{kg}$
• Gravitational constant: $G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$

(a) Magnitude of the gravitational force on the satellite

The gravitational force $F$ is given by Newton's law of gravitation:

$F = \frac{G M_E m}{r^2}$

We know the satellite is at a distance of $r = 2.19 \times R_E$. Substituting this in, we get:

$r = 2.19 \times (6.37 \times 10^6 \, \text{m})$

Now, let's calculate the gravitational force $F$.

(b) Magnitude of the gravitational force exerted on the Earth by the satellite

According to Newton's third law, the gravitational force that the satellite exerts on the Earth is equal in magnitude but opposite in direction to the force the Earth exerts on the satellite. Therefore, the force in part (b) is the same as the force in part (a):

$F_{\text{earth on satellite}} = F_{\text{satellite on earth}} = \text{Result from part (a)}$

(c) Magnitude of the satellite's acceleration

The satellite's acceleration can be found using Newton's second law:

$a_{\text{sat}} = \frac{F}{m}$

We will substitute the force from part (a) and the mass of the satellite to find its acceleration.

(d) Magnitude of the Earth's acceleration

The acceleration of the Earth due to the satellite is also given by Newton's second law:

$a_{\text{earth}} = \frac{F}{M_E}$

Since the Earth's mass is much larger than the satellite's, the acceleration of the Earth will be extremely small.

Let's proceed to calculate each part.### Results: (a) The gravitational force exerted on the 461-kg satellite is approximately $944.15 \, \text{N}$.

(b) The gravitational force exerted on the Earth by the satellite is also $944.15 \, \text{N}$ (same as part (a) due to Newton's third law).

(c) The magnitude of the satellite's acceleration is approximately $2.05 \, \text{m/s}^2$.

(d) The magnitude of the Earth's acceleration due to the satellite is approximately $1.58 \times 10^{-22} \, \text{m/s}^2$. This value is extremely small because of Earth's large mass.

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Would you like more details on any part of the process, or do you have further questions?

Here are 5 related questions to think about:

1. How does the distance between two objects affect the gravitational force between them?
2. What would happen to the gravitational force if the satellite's mass were doubled?
3. Why is the Earth's acceleration so much smaller compared to the satellite's?
4. How would the satellite's acceleration change if it were at a higher orbit?
5. What role does the gravitational constant $G$ play in these calculations?

Tip: Gravitational force decreases with the square of the distance, so even small increases in distance can significantly reduce the force!