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.
Text Solution
Verified by Experts
Here `h=1000 km=1000xx10^(3)m=10^(6)m` `r=R+h=6.38xx10^(6)+10^(6)=7.38xx10^(6)m` Orbital speed, `v_(0)=sqrt((GM)/(R+h))=sqrt((6.67xx10^(-11)xx6xx10^(24))/(7.38xx10^(6)))=7364ms^(-1)` Time period `T=(2pir)/(v_(0))=(2xx(22/7)xx(7.38xx10^(6)))/(7.364xx10^(3))=6297s`
Topper's Solved these Questions
GRAVITATION
CENGAGE PHYSICS|Exercise Exercise 6.1|13 Videos
GRAVITATION
CENGAGE PHYSICS|Exercise Exercise 6.2|15 Videos
FLUID MECHANICS
CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos
KINEMATICS-1
CENGAGE PHYSICS|Exercise Integer|9 Videos
Similar Questions
Explore conceptually related problems
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.
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)
An artificial satellite of mass 200 kg, revolves around the earth in an orbit of average radius 6670 km. What is its orbital kinetic energy ? [M_("earth")=6xx10^(24)kg, G=6.67xx10^(-11)N-m^(2)//kg^(2)]
The distance between centre of the earth and moon is 384000 km . If the mass of the earth is 6xx10^(24) kg and G=6.66xx10^(-11)Nm^(2)//kg^(2) . The speed of the moon is nearly
A rocket is fired vertically upwards with a speed of upsilon (=5 km s^(-1)) from the surface of earth. It goes up to a height h before returning to earth. At height h a body is thrown from the rocket with speed upsilon_(0) in such away so that the body becomes a satellite of earth. Let the mass of the earth, M = 6 xx 10^(24) kg , mean radius of the earth, R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2) . Answer the following questions: Time period of revollution of satellite around the earth is
An artificial satellite is revolving in a circular orbit at height of 1200 km above the surface of the earth. If the radius of the earth is 6400 km and mass is 6xx10^(24) kg , the orbital velocity is
Find the escape velocity of a body from the earth. [R(earth) =6.4xx10^(6)m, rho" (earth)"=5.52xx10^(3)kg//m^(3), G=6.67xx10^(-11)N.m^(2)//kg^(2)]
The radius of the earth is 6.37 xx 10^(6) m and its mean density is 5.5 xx 10^(3) "kg m"^(-3) and G = 6.67 xx 10^(-11) "N-m"^(2) kg^(-2) Find the gravitational potential on the surface of the earth.
