A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-

Related Formula

According to Kepler's Third Law (Law of Periods):

T² ∝ r³$T^2 \propto r^3$

where:

• T$T$ is the time period of revolution.
• r$r$ is the orbital radius of the planet.

Core Logic

Using the proportionality relationship for two states:

Given:

• T₁ = 200 days$T_1 = 200\text{ days}$
• r₂ = (r₁)/(4)$r_2 = \frac{r_1}{4}$

Step 1: Calculate the New Time Period

Substitute the values into the proportionality relation:

Taking the square root on both sides:

Pattern Recognition

If orbital distance scales by x$x$, the period scales by x3/2$x^{3/2}$. Here, distance scales by (1)/(4)$\frac{1}{4}$, so the period scales by ((1)/(4))3/2 = (1)/(8)$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$. Thus, 200 × (1)/(8) = 25 days$200 \times \frac{1}{8} = 25\text{ days}$.

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

Core Logic

Since diameter reduces to half, the radius R₂$R_{2}$ also reduces to half of its initial value R₁$R_{1}$:

Setting up the ratio:

Step 1: Final Calculation

Pattern Recognition

Inverse square dependence means halving the distance scale amplifies the surface field metric by a factor of 2² = 4$2^2 = 4$ matching constant mass bounds.

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

Acceleration due to gravity at a height h$h$ above the Earth's surface:

Acceleration due to gravity at a depth h$h$ below the Earth's surface:

Core Logic

We need the weights to be identical, meaning gₚ = gq$g_p = g_q$ at the exact same distance value h$h$:

Dividing by g$g$ and simplifying the left side denominator fraction:

Step 1: Set Up Algebraic Equation

Let (h)/(R) = x$\frac{h}{R} = x$. Then:

Step 2: Solve for x

Since x ≠ 0$x \neq 0$ (distance cannot be zero), divide by x$x$:

Solving via quadratic formula:

Since distance parameter x gt 0$x \gt 0$, we take the positive root:

Step 3: Find Height h

Substitute back x = (h)/(R)$x = \frac{h}{R}$:

Therefore, the required distance is √(5)R - R2$\frac{\sqrt{5}R - R}{2}$.

Pattern Recognition

Do not use the linear approximation formula gh ≈ g(1 - (2h)/(R))$g_h \approx g(1 - \frac{2h}{R})$ unless the problem explicitly states h ll R$h \ll R$. Equating the approximated form to depth gives hheight = (1)/(2) hdepth$h_{\text{height}} = \frac{1}{2} h_{\text{depth}}$, which fails when looking for a single unified distance value h$h$.

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

Core Logic

For Earth: Vₑ = √((2GME)/(RE)) = 11.2 ~km/s$V_e = \sqrt{\frac{2GM_E}{R_E}} = 11.2 \mathrm{~km/s}$ For the planet: RP = RE3$\mathrm{R}_{\mathrm{P}} = \frac{\mathrm{R}_{\mathrm{E}}}{3}$ and MP = ME6$\mathrm{M}_{\mathrm{P}} = \frac{\mathrm{M}_{\mathrm{E}}}{6}$ We can express the escape velocity of the planet Vₚ$V_p$ as a ratio of the Earth's escape velocity.

Step 1: Ratio of Velocities

Step 2: Calculate Escape Velocity

Pattern Recognition

Any scaling of a planet's mass by factor α$\alpha$ and radius by factor β$\beta$ scales the escape velocity by a factor of √(α / β)$\sqrt{\alpha / \beta}$.

Chapter Mix

Class 11 Physics: Gravitation

Related Formula

Core Logic

The gravitational potential (V$V$) and acceleration due to gravity (g'$g'$) at a distance r = RE + h$r = R_E + h$ from the center of the earth can be related by dividing their magnitudes: |V| / g' = r$|V| / g' = r$.

Step 1: Set Up Equations

From the given data:

Step 2: Isolate Variable

Divide equation (i) by (ii) (taking magnitudes):

Step 3: Solve for h

Given mean radius of the earth RE = 6400 ~km$R_E = 6400 \mathrm{~km}$: 6400 + h = 8000$6400 + h = 8000$

Pattern Recognition

Always exploit the V/g = r$V/g = r$ relationship to extract distances cleanly without having to substitute large values for G$G$ or ME$M_E$.

Chapter Mix

Class 11 Physics: Gravitation