Question

If the orbital speed of a body revolving in a circular path near the surface of the earth is 8 kms$^{-1}$, then the orbital speed of a body revolving around the earth in a circular orbit at height of 19,200 km from the surface of earth is (Radius of the earth = 6400 km)

• A. 4 kms$^{-1}$
• B. 6 kms$^{-1}$
• C. 7.5 kms$^{-1}$
• D. 9 kms$^{-1}$

Hint:

For problems comparing orbital speeds at different altitudes, the proportionality $v_o \propto 1/\sqrt{r}$ (where $r$ is the distance from the planet's center, not the altitude) is the most efficient tool. Set up a ratio to cancel out the constants $G$ and $M$.

Answer 1

The Correct Option is A

The orbital speed ($v_o$) of a satellite revolving around the Earth at a distance $r$ from the center of the Earth is given by the formula:
$v_o = \sqrt{\frac{GM}{r}}$, where G is the gravitational constant and M is the mass of the Earth.
This formula shows that $v_o \propto \frac{1}{\sqrt{r}}$.
Let's consider two cases.
Case 1: Orbit near the surface of the Earth.
The distance from the center is the radius of the Earth, $r_1 = R_E = 6400$ km.
The orbital speed is given as $v_1 = 8$ km/s.
Case 2: Orbit at a height $h = 19200$ km from the surface.
The distance from the center is $r_2 = R_E + h = 6400 \text{ km} + 19200 \text{ km} = 25600$ km.
The orbital speed is $v_2$, which we need to find.
Using the proportionality, we can write a ratio:
$\frac{v_2}{v_1} = \sqrt{\frac{r_1}{r_2}}$.
$\frac{v_2}{8} = \sqrt{\frac{6400}{25600}}$.
$\frac{v_2}{8} = \sqrt{\frac{64}{256}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Solving for $v_2$:
$v_2 = 8 \times \frac{1}{2} = 4$ km/s.