Question 391: Gravitation

Question

A satellite is orbiting just above the surface of the earth with period $$T$$. If $$d$$ is the density of the earth and $$G$$ is the universal constant of gravitation, the quantity $$\frac{3 \pi}{G d}$$ represents :

Accepted Answer

For a satellite orbiting just above the surface of the Earth, we can use the formula for the period T of the satellite in terms of the Earth's density d and the gravitational constant G:
$$T = 2\pi\sqrt{\frac{a^3}{GM}}$$
where a is the semi-major axis of the orbit (which is approximately equal to the Earth's radius R for a satellite orbiting just above the surface) and M is the mass of the Earth.
We can express the mass of the Earth M in terms of its density d and volume:
$$M = dV = d	imes\frac{4}{3}\pi R^3$$
Now, substitute this expression for M into the equation for T:
$$T = 2\pi\sqrt{\frac{a^3}{Gd	imes\frac{4}{3}\pi R^3}}$$
Since the satellite is orbiting just above the Earth's surface, we can approximate a Ã¢ÂÂ R:
$$T = 2\pi\sqrt{\frac{R^3}{Gd	imes\frac{4}{3}\pi R^3}}$$
Simplify the equation:
$$T = 2\pi\sqrt{\frac{1}{Gd	imes\frac{4}{3}\pi}}$$
Now, square both sides of the equation:
$$T^2 = 4\pi^2\frac{1}{Gd	imes\frac{4}{3}\pi}$$
Simplify further:
$$T^2 = \frac{3\pi}{Gd}$$
Thus, the quantity $$\frac{3\pi}{Gd}$$ represents $$T^2$$.

Answer

$$T^{2}$$

Answer

$$T^{3}$$

Answer

$$\sqrt{T}$$

Answer

$$T$$

Gravitation

A satellite is orbiting just above the surface of the earth with period $T. If d is the density of the earth and G is the universal constant of gravitation, the quantity \frac{3 \pi}{G d}$ represents :

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