Home
Class 12
PHYSICS
A particle is launched from a height 8R ...

A particle is launched from a height `8R` above the surface of earth and it is given a speed `((GM)/(8R))` parallel to the surfaces then

A

It will escape fro the gravitational field of earth.

B

It will follow a circular orbit.

C

It will follow an elliptical orbit.

D

the point of launching will be the perigee of its orbit.

Text Solution

AI Generated Solution

To solve the problem, we need to analyze the motion of a particle launched from a height above the Earth's surface with a given speed. Here’s a step-by-step breakdown of the solution: ### Step 1: Understand the Given Parameters - The particle is launched from a height of \( 8R \) above the Earth's surface. - The speed of the particle is given as \( \frac{GM}{8R} \), where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. - The radius of the Earth is \( R \). ### Step 2: Calculate the Escape Velocity ...
Promotional Banner

Similar Questions

Explore conceptually related problems

A satellite ils launched into a circular orbit 1600km above the surface of the earth. Find the period of revolution if the radius of the earth is R=6400km and the acceleration due to gravity is 9.8ms^(-2) . At what height from the ground should it be launched so that it may appear stationary over a point on the earth's equator?

A mass m is taken to a height R from the surface of the earth and then is given a vertical velocity upsilon . Find the minimum value of upsilon , so that mass never returns to the surface of the earth. (Radius of earth is R and mass of the earth m ).

A particle of mass 'm' is raised to a height h = R from the surface of earth. Find increase in potential energy. R = radius of earth. g = acceleration due to gravity on the surface of earth.

A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:

A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:

With what minimum velocity v should a particle be thrown horizontally from a height h above the surface of the earth of mass M so that it does not fall back on the earth ? Radius of earth is R ?

A satellite having time period same as that of the earth's rotation about its own axis is orbiting the earth at a height 8R above the surface of earth. Where R is radius of earth. What will be the time period of another satellite at a height 3.5 R from the surface of earth ?

A projectile is launched from the surface of the earth with a very high speed u at an angle theta with vertical . What is its velocity when it is at the farthest distance from the earth surface . Given that the maximum height reached when it is launched vertically from the earth with a velocity v = sqrt((GM)/(R))

Match the items in column I to the items in column II given below : G = universal gravitational constant, M = mass of earth, R = radius of earth {:(,"Column -I",,"Column -II"),((A),"Escape velocity from a height R above earth's surface",(p),sqrt((4GM)/(R ))),((B),"Minimum horizontal velocity required for a particle at a height R above earth's surface to avoid falling on earth",(q),sqrt((GM)/(R ))),((C ),"Escape valocity on the surface of earth if earth's diameter were shrunk to half of its present value",(r ),sqrt((GM)/(3R))),((D),"Velocity of a body thrown from the earth's surface to reach a height of " (R )/(3) " from the surface of the earth",(s),sqrt((GM)/(2R))):}

A satelite is revolving in a circular orbit at a height h above the surface of the earth of radius R. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. The relation between h and R is