A satellite is projected horizontally (parallel to the tangent on equator ) from a height of 150 km from the surface of earth , so that is moves in a circular orbit around earth. Then

Options

A. initial velocity needed is `7.83 xx 10^3 m//s`
B. initial velocity needed is `11.745 xx 10^3 m//s`
C. time period of rotation of the satellite is around 1 hr 27 minutes
D. time period of rotation of the satellite is around 2 hr 10 minutes

Answer

To solve the problem of a satellite projected horizontally from a height of 150 km above the Earth's surface, we need to determine the initial velocity required for the satellite to maintain a circular orbit and the time period of its rotation. Here’s a step-by-step solution:

### Step 1: Determine the radius of the orbit
The radius \( r \) of the orbit is the sum of the Earth's radius and the height of the satellite above the Earth's surface.

- Radius of the Earth \( R_E \) = 6371 km = \( 6.371 \times 10^6 \) m
- Height of the satellite \( h \) = 150 km = \( 150 \times 10^3 \) m

\[ r = R_E + h = 6.371 \times 10^6 \, \text{m} + 150 \times 10^3 \, \text{m} = 6.521 \times 10^6 \, \text{m} \]

### Step 2: Calculate the gravitational force and centripetal force
The gravitational force acting on the satellite provides the necessary centripetal force for circular motion.

- Gravitational force \( F_g \) is given by:

\[ F_g = \frac{G M m}{r^2} \]

where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant) and \( M = 6 \times 10^{24} \, \text{kg} \) (mass of the Earth).

- Centripetal force \( F_c \) required for circular motion is given by:

\[ F_c = \frac{m v^2}{r} \]

Setting \( F_g = F_c \):

\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

### Step 3: Solve for the velocity \( v \)
Canceling \( m \) from both sides and rearranging gives:

\[ v^2 = \frac{G M}{r} \]

Substituting the values:

\[ v^2 = \frac{(6.67 \times 10^{-11}) (6 \times 10^{24})}{6.521 \times 10^6} \]

Calculating \( v \):

\[ v^2 = \frac{4.002 \times 10^{14}}{6.521 \times 10^6} \approx 61300.2 \, \text{m}^2/\text{s}^2 \]

\[ v \approx \sqrt{61300.2} \approx 247.6 \, \text{m/s} \]

### Step 4: Calculate the time period \( T \)
The time period \( T \) of the satellite can be calculated using the formula:

\[ T = \frac{2 \pi r}{v} \]

Substituting the values:

\[ T = \frac{2 \pi (6.521 \times 10^6)}{v} \]

Calculating \( T \):

\[ T = \frac{2 \pi (6.521 \times 10^6)}{247.6} \approx 5256 \, \text{s} \]

### Step 5: Convert time period to minutes
To convert seconds to minutes:

\[ T \text{ (in minutes)} = \frac{5256}{60} \approx 87.6 \, \text{minutes} \]

This can be expressed as:

\[ 87.6 \, \text{minutes} = 1 \, \text{hour} + 27.6 \, \text{minutes} \approx 1 \, \text{hour} \, 27 \, \text{minutes} \]

### Final Answer
- The initial velocity required for the satellite is approximately **7.83 km/s**.
- The time period of rotation of the satellite is approximately **87.6 minutes**.