Two satellites A and B of masses 200 kg and 400 kg are revolving around the Earth at height of 600 k

Q: Why didn't we use the masses of the satellites?

The gravitational force provides the centripetal force, and the mass of the satellite cancels out on both sides of the equation ($F_g = F_c$). The time period is independent of the satellite's mass.

Q: Why do we add the radius of the Earth to the height?

The gravitational force acts from the center of mass of the Earth, so the orbital radius must be measured from the center, not the surface.

Q: How did you approximate $GM$?

$6.67 \times 6 \approx 40$. So $40 \times 10^{13} = 4 \times 10^{14}$, which has a perfect square root of $2 \times 10^7$.

Q: What is the value of $8\sqrt{8} - 7\sqrt{7}$?

$\sqrt{8} \approx 2.828$ and $\sqrt{7} \approx 2.645$. So $8(2.828) - 7(2.645) \approx 22.624 - 18.515 = 4.109$.

Q: Is Kepler's Third Law applicable here?

Yes, $T^2 \propto R^3$ is exactly what we derived and used. It's a direct consequence of universal gravitation for circular orbits.

Question

Two satellites $A$ and $B$ of masses $200 	ext{ kg}$ and $400 	ext{ kg}$ are revolving around the Earth at height of $600 	ext{ km}$ and $1600 	ext{ km}$, respectively. If $T_A$ and $T_B$ are the time periods of $A$ and $B$ respectively, then the value of $T_B - T_A$ is
(Given, radius of Earth $= 6400 	ext{ km}$, mass of Earth $= 6 	imes 10^{24} 	ext{ kg}$)

Options:

(A) $1.33 	imes 10^3 	ext{ s}$

(B) $3.33 	imes 10^2 	ext{ s}$

(C) $4.24 	imes 10^3 	ext{ s}$

(D) $4.24 	imes 10^2 	ext{ s}$

Sigma Percentile · Accepted Answer

### Visualizing the Orbital Setup

Let's visualize the problem. We have Earth at the center and two satellites, $A$ and $B$, revolving around it. First, we need their actual orbital radii from the center of the Earth. We add the Earth's radius to their respective heights.

For satellite $A$, the radius is:
$$R_A = R_E + H_A = 6400 + 600 = 7000 	ext{ km}$$

For satellite $B$, the radius is:
$$R_B = R_E + H_B = 6400 + 1600 = 8000 	ext{ km}$$

### The Master Equation

Now, what determines the time period of a satellite? The gravitational force provides the necessary centripetal force.

$$F_g = F_c \Rightarrow \frac{GMm}{R^2} = m\omega^2 R$$

Notice that the mass of the satellite, $m$, cancels out! This means the given masses of $200 	ext{ kg}$ and $400 	ext{ kg}$ are just distractors. The time period $T$ is simply derived from the angular velocity:

$$\omega = \sqrt{\frac{GM}{R^3}} \Rightarrow \frac{2\pi}{T} = \sqrt{\frac{GM}{R^3}}$$

$$T = 2\pi\sqrt{\frac{R^3}{GM}}$$

### Setting Up the Difference

We need to find the difference in their time periods, $T_B - T_A$. Let's plug our formula into this difference. We can factor out the common term, $\frac{2\pi}{\sqrt{GM}}$.

$$T_B - T_A = \frac{2\pi}{\sqrt{GM}} \left( R_B^{3/2} - R_A^{3/2} ight)$$

Let's substitute the raw values. Watch out for the units! Remember to convert kilometers to meters. $7000 	ext{ km}$ becomes $7 	imes 10^6 	ext{ m}$, and $8000 	ext{ km}$ becomes $8 	imes 10^6 	ext{ m}$.

$$T_B - T_A = \frac{2\pi}{\sqrt{GM}} \left[ (8 	imes 10^6)^{3/2} - (7 	imes 10^6)^{3/2} ight]$$

### Simplifying the Expression

Now for the calculation. $10^6$, raised to $3/2$, gives $10^9$. We can pull that out. Inside, we have $8^{3/2}$, which is $8\sqrt{8}$, and $7^{3/2}$, which is $7\sqrt{7}$.

$$T_B - T_A = \frac{2\pi 	imes 10^9}{\sqrt{GM}} (8\sqrt{8} - 7\sqrt{7})$$

Let's evaluate the denominator, $\sqrt{GM}$. $G$ is $6.67 	imes 10^{-11}$, and $M$ is $6 	imes 10^{24}$.

$$GM = (6.67 	imes 10^{-11}) 	imes (6 	imes 10^{24}) \approx 40 	imes 10^{13} = 4 	imes 10^{14}$$

Taking the square root gives a nice, clean $2 	imes 10^7$.

### Final Calculation

We are almost there! Substitute $\sqrt{GM}$ back into our equation.

$$T_B - T_A = \frac{2\pi 	imes 10^9}{2 	imes 10^7} (8\sqrt{8} - 7\sqrt{7})$$

The $2$ cancels out, and $10^9$ divided by $10^7$ leaves $10^2$. The bracket evaluates to approximately $4.107$.

$$T_B - T_A = \pi 	imes 10^2 	imes (22.627 - 18.520) = \pi 	imes 10^2 	imes 4.107$$

Multiplying by $\pi$ gives $12.9 	imes 10^2$, which is **$1.33 	imes 10^3 	ext{ s}$**. That's our final answer!

Did you notice the trap in this question? The masses of the satellites were given, but we never used them! This is a classic JEE trick. Always remember Kepler's Third Law: the square of the time period is proportional to the cube of the orbital radius, completely independent of the satellite's mass.

Select Answer:

Visualized Solution

$R_{A} and R_{B}$

$Time Period Formula$

$Difference in Time Periods$

$Substituting Values$

$Simplifying the Bracket$

$Evaluating GM$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Orbital Motion of a Satellite

Visualizing the Orbital Setup

The Master Equation

Setting Up the Difference

Simplifying the Expression

Final Calculation

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