How do you find the time period of a geostationary satellite?

Each satellite has different longitudes and are used for different purposes. Therefore, a period of geostationary satellite is equal to the duration of one whole day on Earth i.e., 23 hours and 56 minutes.

What is the formula of geostationary satellite?

FA=R2GMm where G, equals, 6, point, 67, times, 10, to the power minus 11 , m, cubed, k, g, to the power minus 1 , s, to the power minus 2 ,G=6.67×10−11m3kg−1s−2 is the gravitational constant; M, equals, 5, point, 97, times, 10, to the power 24 , k, g,M=5.97×1024kg and m are the masses of the Earth and the satellite …

What is the time period of geostationary satellite 1?

geostationary orbit, a circular orbit 35,785 km (22,236 miles) above Earth’s Equator in which a satellite’s orbital period is equal to Earth’s rotation period of 23 hours and 56 minutes.

What is the time period and height of geostationary satellite?

Orbital stability A geostationary orbit can be achieved only at an altitude very close to 35,786 kilometres (22,236 miles) and directly above the equator. This equates to an orbital speed of 3.07 kilometres per second (1.91 miles per second) and an orbital period of 1,436 minutes, one sidereal day.

What is the time period?

A time period (denoted by ‘T” ) is the time taken for one complete cycle of vibration to pass a given point. As the frequency of a wave increases, the time period of the wave decreases. The unit for time period is ‘seconds’.

What is the time period and height of geostationary satellite above the surface of Earth?

A satellite placed at a definite height directly above the Earth’s equator and revolves in the same direction as the Earth rotates; so that its orbital time period is same as the Earth’s rotation period (24 hours), is called a Geo-stationary satellite.

What is the time period of revolution of geostationary satellite class 11?

24 hours
The time period of geostationary satellite is 24 hours.

What is the time period of geostationary satellite with respect to earth?

A geostationary orbit can be achieved only at an altitude very close to 35,786 kilometres (22,236 miles) and directly above the equator. This equates to an orbital speed of 3.07 kilometres per second (1.91 miles per second) and an orbital period of 1,436 minutes, one sidereal day.

What is the time period of Pendulum hanged in satellite?

The time period of simple pendulum on the planet, if T is the time period of simple pendulum on earth. Assertion : The time-period of pendulu, on a satellite orbiting the earth is infinity . Reason : Time-period of a pendulum is inversely proportional to √g .

How do you calculate time period and frequency?

Frequency is expressed in Hz (Frequency = cycles/seconds). To calculate the time interval of a known frequency, simply divide 1 by the frequency (e.g. a frequency of 100 Hz has a time interval of 1/(100 Hz) = 0.01 seconds; 500 Hz = 1/(500Hz) = 0.002 seconds, etc.)

How do you calculate time period of a satellite?

Time period of a satellite. Time taken by the satellite to complete one revolution round the Earth is called time period. Time period, T = circumference of the orbit / orbital velocity. T = 2πr / v 0 = 2π(R +h) / v 0. where r is the radius of the orbit which is equal to (R+h).

What is a geostationary satellite?

A geostationary satellite is a satellite in geostationary orbit, with an orbital period the same as the Earth’s rotation period. The geostationary orbit is a circular orbit directly above the Earth’s equator. How high above the Earth’s surface must the geostationary satellite be placed into orbit?

What is the period of time for a satellite to stay stationary?

In cases of stationary satellites, the period, T, is 24 hours, or about 86,400 seconds. Can you find the distance a stationary satellite needs be from the center of the Earth (that is, the radius) to stay stationary?

How do you calculate the radius of a geostationary orbit?

To calculate the radius of a geostationary orbit, the centripetal force must equal the gravitational force on the satellite or mass. We know that (m2) is the mass of the earth at 5.98×10^24 kg, T is the time period and G the universal gravitation constant at 6.67 x10^-11 kg^-2 .