Question

A satellite of mass 500 kg orbits the earth with a period of 6,000 s. The earth has a mass of 5.97 x 10²⁴ kg, a radius of 6.38 x 10⁶ m, and G = 6.67 % 10⁻¹¹ N. m²/kg² a. Calculate the magnitude of the earth's gravitational force on the satellite.

Answer 1

The magnitude of Earth's gravitational force on the satellite can be calculated using Newton's Universal Law of Gravitation formula, involving the gravitational constant, the masses of the Earth and satellite, and the satellite's distance from Earth determined by its orbital period.

Step-by-step explanation:

To calculate the magnitude of the Earth's gravitational force on the satellite, we can use Newton's Universal Law of Gravitation. This law states that the force between two masses is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula for gravitational force (F) is given by:

F = G * (m1 * m2) / r²

Where G is the gravitational constant, m1 is the mass of the Earth, m2 is the mass of the satellite, and r is the distance between the centers of the Earth and the satellite. With the given values:

G = 6.67 x 10⁻¹¹ Nm²/kg²

m1 (Earth's mass) = 5.97 x 10²⁴ kg

m2 (satellite's mass) = 500 kg

r (distance) = Earth's radius + satellite's altitude

We're provided with the period (T) of the satellite, which is 6,000 seconds. From the period, we can find the satellite's orbital radius using the formula for the orbital period of a satellite which relates directly to the distance from the center of the Earth:

T = 2π ∙ √(r³/(G ∙ m1))

After solving for r, we can calculate the gravitational force using the values of G, m1, m2, and r determined from the satellite's period. The force calculated reflects the gravitational pull the Earth exerts on the satellite.

Answer 2

The magnitude of the Earth's gravitational force on the satellite is approximately
`\( 4891.35 \, \text{N} \) (Newtons).`

To calculate the magnitude of the Earth's gravitational force on the satellite, we can use Newton's Law of Universal Gravitation:

`\[ F = G (m_1 m_2)/(r^2) \]`

Where:

-
`\( F \)` is the gravitational force,

-
`\( G \)` is the gravitational constant
`(\(6.67 * 10^(-11) \, \text{N} \cdot \text{m}^2/\text{kg}^2\)),`

-
`\( m_1 \) and \( m_2 \)` are the masses of the two objects (the Earth and the satellite),

-
`\( r \)` is the distance between the centers of the two objects.

Given:

-
`\( m_{\text{satellite}} = 500 \, \text{kg} \),`

-
`\( m_{\text{earth}} = 5.97 * 10^(24) \, \text{kg} \),`

-
`\( G = 6.67 * 10^(-11) \, \text{N} \cdot \text{m}^2/\text{kg}^2 \).`

To find
`\( r \),` the distance between the centers of the Earth and the satellite, we need to add the Earth's radius to the altitude of the satellite. The altitude of the satellite can be determined from its orbital period using Kepler's Third Law, but since the altitude is not directly given, we'll assume for simplicity that the satellite is orbiting close to the Earth's surface (which is a significant approximation).

Therefore,
`\( r \approx \text{Earth's radius} = 6.38 * 10^6 \, \text{m} \).`

Now, we can calculate the gravitational force:

`\[ F = 6.67 * 10^(-11) (500 * 5.97 * 10^(24))/((6.38 * 10^6)^2) \]`

Let's compute this value.

The magnitude of the Earth's gravitational force on the satellite is approximately
`\( 4891.35 \, \text{N} \) (Newtons).`