Q.A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is $1.50 \times 10^{8}\text{ km}$ away from the sun?

The key concept here is that for a circular orbit, the gravitational force provides the necessary centripetal acceleration. The orbital speed depends only on the distance from the center of the Earth, not on the mass of the orbiting body. Near the surface, the orbital radius is essentially the Earth's radius $R$. At a height $h$, the orbital radius becomes $R + h$. Since gravity follows an inverse-square law, the speed scales as the inverse square root of the radius.

Let’s work through it step by step.

1. Write the condition for circular orbit. For a body of mass $m$ orbiting at distance $r$ from Earth’s center, the gravitational force equals the centripetal force:

Cancelling $m$ and simplifying gives:

So orbital speed $v$ is proportional to $1/\sqrt{r}$.

2. Identify the two radii.

   • Near the surface: $r_1 = R = 6400\ \mathrm{km}$, and $v_1 = 8\ \mathrm{km/s}$.
   • At height $h = 19200\ \mathrm{km}$: $r_2 = R + h = 6400 + 19200 = 25600\ \mathrm{km}$.

3. Set up the ratio of speeds.

   From $v^2 \propto 1/r$, we have:

   $$$$

The key concept here is orbital speed in a circular orbit. For a body orbiting a much larger mass (like Earth), the gravitational force provides the necessary centripetal force. The orbital speed $v$ at a distance $r$ from Earth’s center is given by:

where $G$ is the gravitational constant and $M$ is Earth’s mass. This tells us that orbital speed is inversely proportional to the square root of the orbital radius. So if you double the radius, the speed drops by a factor of $\sqrt{2}$; if you quadruple the radius, the speed halves.

Now, let’s work through the problem step by step.

1. Identify the given data

   • Orbital speed near Earth’s surface: $v_1 = 8\ \mathrm{km/s}$
   • Radius of Earth: $R = 6400\ \mathrm{km}$
   • Height above surface: $h = 19,200\ \mathrm{km}$
   • Orbital radius at height $h$: $r_2 = R + h = 6400 + 19,200 = 25,600\ \mathrm{km}$

2. Relate the two orbital speeds

   For a circular orbit near the surface, the orbital radius is essentially $R$ (since the height is negligible compared to Earth’s radius). So:

For the higher orbit:

3. Take the ratio $\frac{v_2}{v_1} = \sqrt{\frac{R}{r_2}}$ …

The key concept here is the radio horizon in line-of-sight (LOS) communication. Because the Earth is curved, even a powerful signal cannot go beyond the tangent line from the antenna to the Earth’s surface. For a single antenna of height $h$, the distance to the horizon (ignoring atmospheric refraction) is $d = \sqrt{2Rh}$, where $R$ is the Earth’s radius. For two antennas, the maximum separation is the sum of their individual horizons — each antenna “sees” the other just above the Earth’s bulge.

Why this works: The geometry is a right triangle with the Earth’s radius as one leg, the line from the antenna to the horizon as the other, and the hypotenuse from the Earth’s center to the antenna tip. Using the Pythagorean theorem and neglecting $h^2$ (since $h \ll R$) gives the simple formula.

Now, step by step:

1. Convert all units to consistent kilometers.

   Earth’s radius $R = 6400$ km.

   Transmitter height $h_t = 33.8$ m $= 0.0338$ km.

   Receiver height $h_r = 64.8$ m $= 0.0648$ km.

2. Write the horizon distance for each antenna.

   For the transmitter: $d_t = \sqrt{2 R h_t}$.

   For the receiver: $d_r = \sqrt{2 R h_r}$.

3. Compute each horizon separately.

   $d_t = \sqrt{2 \times 6400 \times 0.0338} = \sqrt{2 \times 6400 \times 0.0338}$.

   First, $2 \times 6400 = 12800$. Then $12800 \times 0.0338 = 432.64$.

   So $d_t = \sqrt{432.64} = 20.8$ km (since $20.8^2 = 432.64$). …

The key idea is that nuclear matter is nearly incompressible, so the volume of a nucleus is proportional to the number of nucleons. Since volume scales as $R^3$, the radius scales as $A^{1/3}$. This is one of the most reliable empirical facts in nuclear physics — the "nuclear radius formula" $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15}$ m, but here we can find the constant from the given data.

1. Set up the proportionality

   For any nucleus, $R = R_0 A^{1/3}$, where $R_0$ is a constant (the radius of a single nucleon). We are given that for $A_1 = 16$, $R_1 = 3 \times 10^{-15}$ m.

2. Find $R_0$ from the given data

Since $16 = 2^4$, we have $\sqrt[3]{16} = 2^{4/3} = 2 \times 2^{1/3} \approx 2 \times 1.26 = 2.52$, but it's cleaner to keep it symbolic:

3. Apply to the new nucleus For $A_2 = 128$, we have $128 = 2^7$, so $128^{1/3} = 2^{7/3} = 2^2 \times 2^{1/3} = 4 \times 2^{1/3}$. Then:

4. Simplify the ratio $$$$

The core idea here is that the acceleration due to gravity at a height $h$ above Earth’s surface falls off as the inverse square of the distance from Earth’s centre. The given fraction $\frac{16}{49} g_0$ tells us exactly how far the satellite is from the centre. Once we know that distance, Kepler’s third law (which is just Newton’s law of gravitation combined with circular motion) gives the time period directly. The problem then asks us to match that period to the form $K \left[ \frac{\pi^2 R^3}{GM} \right]$, so we solve for $K$.

1. Relate gravity at height to gravity on surface On Earth’s surface: $g_0 = \frac{GM}{R^2}$. At a height $h$ above the surface, distance from centre is $r = R + h$, and

Cancel $GM$ from both sides:

So $r = \frac{7}{4} R$ (taking the positive root). The satellite orbits at a distance $r = \frac{7}{4}R$ from Earth’s centre.

2. Apply Kepler’s third law for a circular orbit For a satellite in a circular orbit, the centripetal force is provided by gravity:

where $\omega = \frac{2\pi}{T}$. This gives

This is Kepler’s third law in Newton’s form.

3. Substitute $r = \frac{7}{4}R$

Simplify:

4. Compare with the given form The problem states $T^2 = K \left[ \frac{\pi^2 R^3}{GM} \right]$. From our result, $K = \frac{343}{16}$. …