Edits to “Statement”, “Solution”, “Answer”

Tete edited Aug 7, 2026, 3:47 AM

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revision #19871 parent #19870 ← older

en/13.4.8.md +3 −3

		@@ -1,15 +1,15 @@
			### Statement

		−	$13.4.8.$ [Insert the problem statement]
		+	$13.4.8.$ An artificial satellite of the Earth with a radius of $1$ m, illuminated by the Sun, is visible from a distance of $300$ km as an ordinary star. Estimate the distance to such a star.

			### Solution

			Let $I$ be the intensity of light from the Sun on the Earth and the satellite. Consequently, the power of light from the Sun intercepted by the cross section of the satellite is $I\cdot\pi r^2$, where $r$ is the radius of the satellite. Since the light is reflected by a half of the satellite, the intensity of light from the satellite on the Earth is

			\[\frac{I\cdot\pi r^2}{2\pi d^2}=I\cdot\frac{1}{2}\left(\frac{r}{d}\right)^2,\]

		−	where $d$ is the distance between the satellite and the Earth. Using $r=1$ m and $d=300$ km, we see that the intensity of the light on the Earth from the satellite is $1/(18\times10^{10})$ of that from the Sun. Since light intensity varies inversely with the square of the distance from the source, a star comparable to the Sun will give the same intensity of light on the Earth if it is $\sqrt(18\times10^{10})\approx4.2\times10^5$ times as far from the Earth as the Sun. As the Sun is about $500$ light-second from the Earth, the star is about $500\times4.2\times10^5$ light-second, which is roughly $6.6$ light-year, from the Earth. Note that the Sun as a yellow dwarf is in fact brighter than most stars which are dwarfs, so a better estimate of the distance is less than $6.6$ light-year.
		+	where $d$ is the distance between the satellite and the Earth. Using $r=1$ m and $d=300$ km, we see that the intensity of the light on the Earth from the satellite is $1/(18\times10^{10})$ of that from the Sun. Since light intensity varies inversely with the square of the distance from the source, a star comparable to the Sun will give the same intensity of light on the Earth as the satellite if it is $\sqrt{18\times10^{10}}\approx4.2\times10^5$ times as far from the Earth as the Sun. Because the Sun is about $500$ light-second from the Earth, the star is about $500\times4.2\times10^5$ light-seconds, which is roughly $6.6$ light-years, from the Earth. Note that the Sun as a yellow dwarf is in fact brighter than most stars which are red dwarfs, so a better estimate of the distance is less than $6.6$ light-year.

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	Less than $6.6$ light-years.

	@@ -1,15 +1,15 @@
	### Statement		### Statement

	$13.4.8.$ [Insert the problem statement]		$13.4.8.$ An artificial satellite of the Earth with a radius of $1$ m, illuminated by the Sun, is visible from a distance of $300$ km as an ordinary star. Estimate the distance to such a star.

	### Solution		### Solution

	Let $I$ be the intensity of light from the Sun on the Earth and the satellite. Consequently, the power of light from the Sun intercepted by the cross section of the satellite is $I\cdot\pi r^2$, where $r$ is the radius of the satellite. Since the light is reflected by a half of the satellite, the intensity of light from the satellite on the Earth is		Let $I$ be the intensity of light from the Sun on the Earth and the satellite. Consequently, the power of light from the Sun intercepted by the cross section of the satellite is $I\cdot\pi r^2$, where $r$ is the radius of the satellite. Since the light is reflected by a half of the satellite, the intensity of light from the satellite on the Earth is

	\[\frac{I\cdot\pi r^2}{2\pi d^2}=I\cdot\frac{1}{2}\left(\frac{r}{d}\right)^2,\]		\[\frac{I\cdot\pi r^2}{2\pi d^2}=I\cdot\frac{1}{2}\left(\frac{r}{d}\right)^2,\]

	where $d$ is the distance between the satellite and the Earth. Using $r=1$ m and $d=300$ km, we see that the intensity of the light on the Earth from the satellite is $1/(18\times10^{10})$ of that from the Sun. Since light intensity varies inversely with the square of the distance from the source, a star comparable to the Sun will give the same intensity of light on the Earth if it is $\sqrt(18\times10^{10})\approx4.2\times10^5$ times as far from the Earth as the Sun. As the Sun is about $500$ light-second from the Earth, the star is about $500\times4.2\times10^5$ light-second, which is roughly $6.6$ light-year, from the Earth. Note that the Sun as a yellow dwarf is in fact brighter than most stars which are dwarfs, so a better estimate of the distance is less than $6.6$ light-year.		where $d$ is the distance between the satellite and the Earth. Using $r=1$ m and $d=300$ km, we see that the intensity of the light on the Earth from the satellite is $1/(18\times10^{10})$ of that from the Sun. Since light intensity varies inversely with the square of the distance from the source, a star comparable to the Sun will give the same intensity of light on the Earth as the satellite if it is $\sqrt{18\times10^{10}}\approx4.2\times10^5$ times as far from the Earth as the Sun. Because the Sun is about $500$ light-second from the Earth, the star is about $500\times4.2\times10^5$ light-seconds, which is roughly $6.6$ light-years, from the Earth. Note that the Sun as a yellow dwarf is in fact brighter than most stars which are red dwarfs, so a better estimate of the distance is less than $6.6$ light-year.

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		Less than $6.6$ light-years.