A satellite is revolving round the earth in circular orbit

Options

A. if mass of earth is made four times, keeping other factors constant, orbital speed of satellite will become two times
B. corresponding to change in part (a), times period of satellite will remain half
C. when value of G is made two times orbital speed increases and time period decreases
D. G has no effect on orbital speed and time period

Explanation

To solve the problem regarding a satellite revolving around the Earth in a circular orbit, we will follow these steps:

### Step 1: Understand the Forces Acting on the Satellite
A satellite in a circular orbit experiences gravitational force acting as the centripetal force. The gravitational force can be expressed as:
\[ F_g = \frac{G M m}{r^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.

### Step 2: Set Up the Equation for Circular Motion
The centripetal force required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where:
- \( v \) is the orbital speed of the satellite.

### Step 3: Equate Gravitational Force and Centripetal Force
For the satellite to remain in orbit, the gravitational force must equal the centripetal force:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{G M}{r} = v^2 \]
Thus, we can express the orbital speed \( v \) as:
\[ v = \sqrt{\frac{G M}{r}} \]

### Step 5: Determine the Time Period of the Satellite
The time period \( T \) of the satellite is the time taken to complete one full orbit. It can be calculated as:
\[ T = \frac{2 \pi r}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2 \pi r}{\sqrt{\frac{G M}{r}}} \]
This simplifies to:
\[ T = 2 \pi \sqrt{\frac{r^3}{G M}} \]

### Step 6: Analyze Changes in Mass of the Earth
If the mass of the Earth is increased to \( 4M \):
1. The new orbital speed \( v' \) becomes:
\[ v' = \sqrt{\frac{G (4M)}{r}} = 2 \sqrt{\frac{G M}{r}} = 2v \]
Thus, the orbital speed doubles.

2. The new time period \( T' \):
\[ T' = 2 \pi \sqrt{\frac{r^3}{G (4M)}} = \frac{1}{2} \cdot 2 \pi \sqrt{\frac{r^3}{G M}} = \frac{T}{2} \]
Thus, the time period is halved.

### Step 7: Analyze Changes in Gravitational Acceleration
If the value of \( g \) (acceleration due to gravity) is doubled:
1. The orbital speed \( v \) increases since \( v \) is directly proportional to \( g \).
2. The time period \( T \) decreases since it is inversely proportional to the square root of \( g \).

### Conclusion
From the analysis:
- If the mass of the Earth is quadrupled, the orbital speed doubles and the time period halves.
- If \( g \) is doubled, the orbital speed increases, and the time period decreases.