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A remote-sensing satellite of earth revo...

A remote-sensing satellite of earth revolves in a circular orbit at a height of `0.25xx10^(6)m` above the surface of earth. If earth's radius is `6.38xx10^(6)m` and `g=9.8ms^(-2)`, then the orbital speed of the satellite is

A

`7.76 kms^(-1)`

B

`8.56 kms^(-1)`

C

`9.13 kms^(-1)`

D

`6.67 kms^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the orbital speed of a satellite revolving in a circular orbit above the Earth's surface, we can follow these steps: ### Step 1: Identify the given values - Height of the satellite above the Earth's surface, \( h = 0.25 \times 10^6 \, m \) - Radius of the Earth, \( R = 6.38 \times 10^6 \, m \) - Acceleration due to gravity at the surface of the Earth, \( g = 9.8 \, m/s^2 \) ### Step 2: Calculate the distance from the center of the Earth to the satellite The total distance \( r \) from the center of the Earth to the satellite is given by: \[ r = R + h \] Substituting the values: \[ r = 6.38 \times 10^6 \, m + 0.25 \times 10^6 \, m = 6.63 \times 10^6 \, m \] ### Step 3: Use the formula for orbital speed The formula for the orbital speed \( v \) of a satellite in a circular orbit is given by: \[ v = \sqrt{\frac{GM}{r}} \] Where \( G \) is the gravitational constant. However, we can also express \( GM \) in terms of \( g \) and \( R \): \[ GM = gR^2 \] Thus, the orbital speed can be rewritten as: \[ v = \sqrt{\frac{gR^2}{r}} \] ### Step 4: Substitute the known values into the formula Now, substituting \( g = 9.8 \, m/s^2 \), \( R = 6.38 \times 10^6 \, m \), and \( r = 6.63 \times 10^6 \, m \): \[ v = \sqrt{\frac{9.8 \times (6.38 \times 10^6)^2}{6.63 \times 10^6}} \] ### Step 5: Calculate the value First, calculate \( (6.38 \times 10^6)^2 \): \[ (6.38 \times 10^6)^2 = 4.065044 \times 10^{13} \, m^2 \] Now substitute this back into the equation: \[ v = \sqrt{\frac{9.8 \times 4.065044 \times 10^{13}}{6.63 \times 10^6}} \] Calculating the numerator: \[ 9.8 \times 4.065044 \times 10^{13} \approx 3.98 \times 10^{14} \] Now divide by \( 6.63 \times 10^6 \): \[ \frac{3.98 \times 10^{14}}{6.63 \times 10^6} \approx 5.99 \times 10^7 \] Finally, take the square root: \[ v \approx \sqrt{5.99 \times 10^7} \approx 7.76 \times 10^3 \, m/s \] ### Step 6: Convert to kilometers per second To convert \( m/s \) to \( km/s \): \[ v \approx 7.76 \, km/s \] ### Final Answer The orbital speed of the satellite is approximately \( 7.76 \, km/s \). ---
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