Home
Class 11
PHYSICS
An artificial satellite circles around t...

An artificial satellite circles around the earth at a height of 2,200 km. Calculate its orbital velocity and period of revolution. Take,
Radius of earth `= 6.37 xx 10^(3) km`
Mass of earth `= 6 xx 10^(24) kg`
`G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`

Text Solution

AI Generated Solution

To solve the problem of calculating the orbital velocity and period of revolution of an artificial satellite at a height of 2,200 km above the Earth's surface, we can follow these steps: ### Step 1: Calculate the Radius of the Orbit The radius \( R \) of the satellite's orbit is the sum of the Earth's radius and the height of the satellite above the Earth's surface. Given: - Radius of the Earth \( R_E = 6.37 \times 10^3 \) km - Height of the satellite \( h = 2,200 \) km ...
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    MODERN PUBLICATION|Exercise Practice Problems|31 Videos

  • GRAVITATION

    MODERN PUBLICATION|Exercise Conceptual Questions|19 Videos

  • MATHEMATICAL TOOLS

    MODERN PUBLICATION|Exercise PRACTICE PROBLEMS (10)|12 Videos

Similar Questions

Explore conceptually related problems

An artificial satellite circled around the earth at a distance of 3400 km. Calculate its orbital velocity and period of revolution. Radius of earth =6400 km and g=9.8 ms^(-2) .

A satellite is revolving round the earth at a height of 600 km. find a. the speed f the satelite and b. the time period of thesatellit. Radius of the earth =6400 km and mass of the earth =6xx10^24kg .

Calculate the angular momentum of earth rotating about its own axis. Take Radius of earth =6.4xx10^(6)m Mass of earth =5.97xx10^(24)kg

Find the escape velocity of a body from the earth. [M (earth ) = 6 xx 10^(24) kg, R ( earth) = 6.4 xx 10^(6) m. G = 6.67 xx 10^(-11) N.m^(2)//kg^(2)]

Calculate (i) kinetic energy (ii) potential energy and (iii) total energy of a satellite of mass 200 kg orbiting around the earth in an orbit of height 100 km from the surface of earth. Given, mass of earth = 10^(25) kg ,radius of earth = 6.4 xx 10^(6) m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

An artificial satellite revolves round the earth at a height of 1000 km . The radius of the earth is 6.38xx10^(3)km . Mass of the earth 6xx10^(24)kg,G=6.67xx10^(-11)Nm^(2)kg^(-2) . Find the orbital speed and period of revolution of the satellite.

A satellite revolves around the earth at a height of 1000 km. The radius of the earth is 6.38 xx 10^(3) km. Mass of the earth is 6xx10^(24)kg and G=6.67xx10^(-14)"N-m"^(2)kg^(-2) . Determine its orbital velocity and period of revolution.

What is the force of gravity on a body of mass 150 kg lying on earth (Mass of earth =6 xx 10^(24) kg Radius of earth = 6 .4 xx 10^(6) m G= 6 .7 xx 10-11 Nm^2//kg^2)

A mango of mass 0.3 kg falls from a tree. Calculate the acceleration of the mango towards the earth. Also calculate the acceleration of the earth towards the mango. Take, Mass of the earth = 5.983 xx 10^(24) kg Radius of the earth = 6.378 xx 10^(6) m G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)

An apple of mass 0.25 kg falls from a tree . What is the acceleration of the apple towards the earth ? Also calculate the acceleration of the earth towards the apple . Mass of the earth = 5.983 xx 10^(24) kg , Radius of the earth = 6.378 xx 10^(6) m and G = 6.67 xx 10^(-11)Nm^(2) kg^(-2)