Gravitational Force Calculation

This problem requires calculating the gravitational force between the Earth and a satellite orbiting it. We will use Newton's Law of Universal Gravitation. Let's break down the steps and calculations involved.

Physics Values Provided

First, let's list all the known values provided in the question:

• Mass of the Earth ($M_E$): $6 \times 10^{24}$ kg
• Mass of the satellite ($m_s$): 400 kg
• Altitude of the satellite from Earth's surface: 400 km
• Radius of the Earth ($R_E$): 6400 km
• Universal Gravitational Constant ($G$): $6.67 \times 10^{-11}$ N m$^2$/kg$^2$

Orbital Distance Calculation

Newton's Law of Gravitation uses the distance between the centers of the two masses. Therefore, we need to calculate the total distance ($r$) from the center of the Earth to the satellite. This is the sum of the Earth's radius and the satellite's altitude.

Before adding, we must convert the distances from kilometers (km) to meters (m), as the gravitational constant ($G$) is given in terms of meters.

• Altitude in meters: $400 \text{ km} = 400 \times 1000 \text{ m} = 0.4 \times 10^6 \text{ m}$
• Earth's Radius in meters: $R_E = 6400 \text{ km} = 6400 \times 1000 \text{ m} = 6.4 \times 10^6 \text{ m}$

Now, calculate the total orbital radius ($r$):

$r = R_E + \text{Altitude}$
$r = (6.4 \times 10^6 \text{ m}) + (0.4 \times 10^6 \text{ m})$
$r = 6.8 \times 10^6 \text{ m}$

Applying Newton's Law

The formula for the gravitational force ($F$) between two masses ($M_1$ and $M_2$) separated by a distance ($r$) is:

$F = G \frac{M_1 M_2}{r^2}$

In our case, $M_1$ is the mass of the Earth ($M_E$) and $M_2$ is the mass of the satellite ($m_s$).

Detailed Force Calculation

Let's substitute the values into the formula:

$F = (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times \frac{(6 \times 10^{24} \text{ kg}) \times (400 \text{ kg})}{(6.8 \times 10^6 \text{ m})^2}$

First, calculate the product of the masses:

$M_E \times m_s = (6 \times 10^{24} \text{ kg}) \times (400 \text{ kg}) = 2400 \times 10^{24} \text{ kg}^2 = 2.4 \times 10^{27} \text{ kg}^2$

Next, calculate the square of the orbital radius:

$r^2 = (6.8 \times 10^6 \text{ m})^2 = (6.8)^2 \times (10^6)^2 \text{ m}^2 = 46.24 \times 10^{12} \text{ m}^2$

Now, plug these values back into the force equation:

$F = (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times \frac{2.4 \times 10^{27} \text{ kg}^2}{46.24 \times 10^{12} \text{ m}^2}$

Simplify the fraction:

$\frac{2.4 \times 10^{27}}{46.24 \times 10^{12}} = \frac{2.4}{46.24} \times \frac{10^{27}}{10^{12}} \approx 0.05189 \times 10^{(27-12)} = 0.05189 \times 10^{15}$

Now, multiply by G:

$F \approx (6.67 \times 10^{-11}) \times (0.05189 \times 10^{15})$ N

$F \approx (6.67 \times 0.05189) \times (10^{-11} \times 10^{15})$ N

$F \approx 0.34614 \times 10^{(15-11)}$ N

$F \approx 0.34614 \times 10^4$ N

Finally, express the answer in standard notation:

$F \approx 3461.4$ N

Final Force Result

The calculated force exerted by the Earth on the satellite is approximately 3461.4 N.