Two satellites P and Q are moving in different circular orbits around the Earth (radius R). The heig

Q: Why do we use $R+h$ instead of just $h$ in the formula?

Gravity acts from the center of mass of the Earth. The distance $r$ in Newton's law of gravitation is the distance between the centers of the two bodies, which is the Earth's radius plus the height above the surface.

Q: Can I just assume $g$ is proportional to $1/h^2$?

No, that's a very common trap! $g$ is proportional to $1/r^2$, where $r = R+h$. If $h$ is very small compared to $R$, we can use approximations, but here $h$ is comparable to $R$ ($R/3$), so we must use the exact formula.

Q: What happened to the mass of the satellites?

The acceleration due to gravity $g$ experienced by an object is independent of its own mass. It only depends on the mass of the Earth $M$ and the distance $r$.

Q: Why did we take the positive square root of $36/25$?

Because distances ($r_P$ and $r_Q$) are physical quantities that must be positive. A negative ratio wouldn't make physical sense here.

Q: Is this $g$ the same as $9.8 m/s^2$?

No, $9.8 m/s^2$ is the value of $g$ at the surface of the Earth (where $r=R$). As you go higher, $r$ increases, and $g$ decreases.

Question

Two satellites P and Q are moving in different circular orbits around the Earth (radius R). The heights of P and Q from the Earth surface are $h_P$ and $h_Q$, respectively, where $h_P = R/3$. The accelerations of P and Q due to Earth's gravity are $g_P$ and $g_Q$, respectively. If $g_P / g_Q = 36/25$, what is the value of $h_Q$?

Options:

(A) $\frac{3R}{5}$

(B) $\frac{R}{6}$

(C) $\frac{6R}{5}$

(D) $\frac{5R}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at the Earth from space. We have two satellites, P and Q, orbiting at different heights. The problem gives us the height of satellite P and the ratio of the gravitational accelerations they experience. Our goal is to find the height of satellite Q.

First, let's establish the distances of these satellites from the center of the Earth. The distance $r$ is simply the radius of the Earth $R$ plus the height $h$ above the surface. So, for satellite P, $r_P = R + h_P$, and for satellite Q, $r_Q = R + h_Q$.

### The Master Equation

The acceleration due to gravity at a distance $r$ from the center of the Earth is given by Newton's law of universal gravitation:

$$g = \frac{GM}{r^2}$$

This tells us that the acceleration is inversely proportional to the square of the distance from the center. Let's write this for both satellites:

$$g_P = \frac{GM}{r_P^2}$$

$$g_Q = \frac{GM}{r_Q^2}$$

### Finding the Ratio of Radii

We are given the ratio of their accelerations, $g_P / g_Q = 36/25$. Let's divide our two equations. The $GM$ terms cancel out beautifully:

$$\frac{g_P}{g_Q} = \left(\frac{r_Q}{r_P}\right)^2$$

Substituting the given value:

$$\frac{36}{25} = \left(\frac{r_Q}{r_P}\right)^2$$

Taking the square root of both sides, we get a clean ratio for their distances:

$$\frac{r_Q}{r_P} = \frac{6}{5}$$

### Final Calculation

Now, let's find the actual distance of satellite P from the center. We know its height $h_P = R/3$.

$$r_P = R + \frac{R}{3} = \frac{4R}{3}$$

Using the ratio we just found, we can calculate $r_Q$:

$$r_Q = \frac{6}{5} r_P = \frac{6}{5} \left(\frac{4R}{3}\right) = \frac{8R}{5}$$

Finally, to find the height of satellite Q, we subtract the Earth's radius $R$ from $r_Q$:

$$h_Q = r_Q - R = \frac{8R}{5} - R = \frac{3R}{5}$$

And there we have it! The height of satellite Q is **$3R/5$**.

Select Answer:

Visualized Solution

$r = R + h$

$g = \frac{GM}{r ^{2}}$

$\frac{g _{P}}{g _{Q}} = (\frac{r _{Q}}{r _{P}})^{2}$

$\frac{r _{Q}}{r _{P}} = \frac{6}{5}$

$r_{P} = \frac{4 R}{3}$

$r_{Q} = \frac{8 R}{5}$

$h_{Q} = \frac{3 R}{5}$

$v = \frac{GM}{r}$

The Sigma Insight: Acceleration due to Gravity and its Variation

Analyzing the Setup

The Master Equation

Finding the Ratio of Radii

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=R+h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g=r2GM​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } gQ​gP​​=(rP​rQ​​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rP​rQ​​=56​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rP​=34R​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rQ​=58R​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } hQ​=53R​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=rGM​​