Answer :
Sure, let's solve this step-by-step. Given data:
- Gravitational constant, [tex]\( G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} \)[/tex]
- Radius of the Earth, [tex]\( R = 6.4 \times 10^6 \, \text{m} \)[/tex]
- Mass of the Earth, [tex]\( M_e = 6 \times 10^{24} \, \text{kg} \)[/tex]
- Height of the satellite above the Earth's surface, [tex]\( h = 35900 \, \text{km} = 35900 \times 10^3 \, \text{m} \)[/tex]
### Step 1: Calculate the total distance from the center of the Earth to the satellite.
The total distance [tex]\( r \)[/tex] from the center of the Earth to the satellite is given by the sum of the Earth's radius [tex]\( R \)[/tex] and the height of the satellite [tex]\( h \)[/tex]:
[tex]\[ r = R + h \][/tex]
Substitute the given values:
[tex]\[ r = 6.4 \times 10^6 \, \text{m} + 35900 \times 10^3 \, \text{m} = 42300 \times 10^3 \, \text{m} = 4.23 \times 10^7 \, \text{m} \][/tex]
### Step 2: Calculate the orbital speed of the satellite.
To calculate the orbital speed [tex]\( v \)[/tex], we use the formula given by:
[tex]\[ v = \sqrt{\frac{GM_e}{r}} \][/tex]
Substitute the given values for [tex]\( G \)[/tex], [tex]\( M_e \)[/tex], and [tex]\( r \)[/tex]:
[tex]\[ v = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{4.23 \times 10^7}} \][/tex]
Upon calculation, we find the orbital speed [tex]\( v \)[/tex]:
[tex]\[ v \approx 3075.87 \, \text{m/s} \][/tex]
### Step 3: Calculate the orbital period of the satellite.
The orbital period [tex]\( T \)[/tex] is the time it takes for the satellite to complete one full orbit around the Earth. It is given by the formula:
[tex]\[ T = \frac{2 \pi r}{v} \][/tex]
Substitute the values for [tex]\( r \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ T = \frac{2 \pi \times 4.23 \times 10^7 \, \text{m}}{3075.87 \, \text{m/s}} \][/tex]
Upon calculation, we find the orbital period [tex]\( T \)[/tex]:
[tex]\[ T \approx 86407.59 \, \text{s} \][/tex]
### Final Results:
1. The orbital speed of the satellite is approximately [tex]\( v \approx 3075.87 \, \text{m/s} \)[/tex].
2. The period of revolution (orbital period) of the satellite is approximately [tex]\( T \approx 86407.59 \, \text{s} \)[/tex].
These results represent the speed and the time it takes for the satellite to complete one full orbit around the Earth at the given altitude.
- Gravitational constant, [tex]\( G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} \)[/tex]
- Radius of the Earth, [tex]\( R = 6.4 \times 10^6 \, \text{m} \)[/tex]
- Mass of the Earth, [tex]\( M_e = 6 \times 10^{24} \, \text{kg} \)[/tex]
- Height of the satellite above the Earth's surface, [tex]\( h = 35900 \, \text{km} = 35900 \times 10^3 \, \text{m} \)[/tex]
### Step 1: Calculate the total distance from the center of the Earth to the satellite.
The total distance [tex]\( r \)[/tex] from the center of the Earth to the satellite is given by the sum of the Earth's radius [tex]\( R \)[/tex] and the height of the satellite [tex]\( h \)[/tex]:
[tex]\[ r = R + h \][/tex]
Substitute the given values:
[tex]\[ r = 6.4 \times 10^6 \, \text{m} + 35900 \times 10^3 \, \text{m} = 42300 \times 10^3 \, \text{m} = 4.23 \times 10^7 \, \text{m} \][/tex]
### Step 2: Calculate the orbital speed of the satellite.
To calculate the orbital speed [tex]\( v \)[/tex], we use the formula given by:
[tex]\[ v = \sqrt{\frac{GM_e}{r}} \][/tex]
Substitute the given values for [tex]\( G \)[/tex], [tex]\( M_e \)[/tex], and [tex]\( r \)[/tex]:
[tex]\[ v = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{4.23 \times 10^7}} \][/tex]
Upon calculation, we find the orbital speed [tex]\( v \)[/tex]:
[tex]\[ v \approx 3075.87 \, \text{m/s} \][/tex]
### Step 3: Calculate the orbital period of the satellite.
The orbital period [tex]\( T \)[/tex] is the time it takes for the satellite to complete one full orbit around the Earth. It is given by the formula:
[tex]\[ T = \frac{2 \pi r}{v} \][/tex]
Substitute the values for [tex]\( r \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ T = \frac{2 \pi \times 4.23 \times 10^7 \, \text{m}}{3075.87 \, \text{m/s}} \][/tex]
Upon calculation, we find the orbital period [tex]\( T \)[/tex]:
[tex]\[ T \approx 86407.59 \, \text{s} \][/tex]
### Final Results:
1. The orbital speed of the satellite is approximately [tex]\( v \approx 3075.87 \, \text{m/s} \)[/tex].
2. The period of revolution (orbital period) of the satellite is approximately [tex]\( T \approx 86407.59 \, \text{s} \)[/tex].
These results represent the speed and the time it takes for the satellite to complete one full orbit around the Earth at the given altitude.
