Physics, asked by HuzaifaMari, 11 months ago

7. A communications satellite which takes 24 hours to orbit the Earth is replaced by a new
satellite which has twice the mass of the first. If the new satellite also has an orbit time of
24hours, then the ratio radius of orbit of new satellite is
radius of orbit of old satellite
(b) 1​

Answers

Answered by jay727
1

Since they match the rotation rate of the Earth (they are geosynchronous), these satellites appear to remain fixed in the sky when viewed from the Earth (they are geostationary).

There are two major types of satellite that can be found in the geosynchronous/geostationary earth orbit (GEO).

Telecommunications Satellites: Since they maintain a fixed position in the sky, a fixed antenna can be used relay messages between the ground and a GEO satellite. Point the antenna in the right direction once and your good to go for the lifetime of the satellite or the antenna (whichever comes first). Non GEO satellites move across the sky and must be tracked, which means building a steering mechanism for the antenna and writing computer software to drive it. GEO satellites also provide 24 hour coverage to large sections of the planet, which makes them great for broadcasting to an information-hungry world that never sleeps. Finally, a set of at least three geostationary satellites spaced evenly around the equator could be used as a relay. Just equip each satellite with one antenna pointing "forward" to the satellite ahead, one pointing "backward" to the satellite behind, and one pointing downward to the Earth. This makes it possible to send wireless messages (audio, video, text, etc.) from nearly anywhere on the Earth to nearly everywhere on the Earth. Because of their great height above the Earth, signals sent to and from GEO satellites are delayed by about a half a second each way or about a whole second round trip. This delay is known as latency and is annoying during a phone call. Annoying latency is getting to be a thing of the past now that most international calls are handled by fiber optic cables laying near the surface of the Earth and true satellite phones have been developed with omnidirectional antennas capable of accessing satellites in low earth orbit (LEO).

Meteorological Satellites: Satellites in geostationary orbits stay fixed over a point on the surface of the Earth and are high enough that they can monitor the weather over nearly an entire continent or ocean. Images taken from the same location in space are easier to use than images taken from a location that changes. Images taken from weather satellites in lower orbits must be processed to correct for changes in perspective and scale and then mosaicked or stitched together. As with telecommunication, weather coverage is a 24 hour a day need in the Twenty-first Century. (Weather never sleeps, too.) Height has its drawbacks as well as its benefits. Being so far up means a loss in resolution. The information you get from a GEO satellite image may not be finely detailed enough. This is why government weather services now operate weather satellites in low earth orbit (LEO) as well as geostationary earth orbit (GEO).

Start with the basic principle behind all circular orbits. The gravitational force is the centripetal force.

Fg = Fc

Fg = Fc

Fg = Fc Gm1m2 = mv2

Fg = Fc Gm1m2 = mv2 r2 r

Replace speed with circumference divided by the period.

Gm1m2 = m2πr ⎞2

r2 r T

Gm = 4π2r

r2 T2

Solve for radius to arrive at a general equation

r = ⎛⎝ GmT2 ⎞⅓⎠

4π2

This problem can also be solved using Kepler's third law of planetary motion: the square of the period of a satellite in a circular orbit is proportional to the cube of its radius. That solution is presented in an earlier section of this book.

Now substitute in the appropriate values. The period of the Earth's rotation is approximately equal to the mean solar day (24 × 3600 s = 86,400 s), but for best results use the sidereal day (86,164 s).

⬇️(to the power 11)

r = (6.67 × 10−11 Nm2/kg2)

(5.97 ×1024 kg)(86,164 s)2 (to the power)

4π2

r = 4.215 × 107 m

r = 42,150 km

Convert to earth radii.

(⬇️to the power 7)

r = 4.215 × 10 7 m

6,371,000 m

r = 6.616 earth radii

r ≈ 7 earth radii

Convert to earth-moon distances.

r = 4.215 × 10 7 m

384,400,000 m

r = 0.1097 earth-moon distance

r ≈ ⅑ earth-moon distance

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