If due to air drag, the orbital radius of satelite decreases from R to `R-DeltaR,DeltaRltltR`, then the expression for change is orbital velocity `Deltav` is [mass of earth M]:-
A planet is revolving around the Sun in an elliptical orbit. Its closest distance from the sun is r_(min) . The farthest distance from the sun is r_(max) if the orbital angular velocity of the planet when it is nearest to the Sun omega then the orbital angular velocity at the point when it i at the farthest distanec from the sun is
A satellite of earth of mass 'm' is taken from orbital radius 2R to 3R, then minimum work done is :-
A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R, h ltlt R). The minimum increase in its orbital velocity required, So that the satellite could escape from the erth's gravitational field, is close to :(Neglect the effect of atomsphere.)
If g is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is
Obtain an expression of acceleration produced by gravity of earth. OR Obtain general equation of gravitation force at distance r from the centre of earth and derive the equation of acceleration due to gravity on the surface of earth.
A spaceship is launched in to a circular orbit of radius R close to surface of earth. The additional velocity to be imparted to the spaceship in the orbit to overcome the earth's gravitational pull is ( g =acceleration due to gravity)
A satellite of mass m is in an elliptical orbit around the earth of mass M(M gt gt m) the speed of the satellite at its nearest point ot the earth (perigee) is sqrt((6GM)/(5R)) where R= its closest distance to the earth it is desired to transfer this satellite into a circular orbit around the earth of radius equal its largest distance from the earth. Find the increase in its speed to be imparted at the apogee (farthest point on the elliptical orbit).
The minimum and maximum distances of a satellite from the centre of the earth are 2R and 4R respectively where R is the radius of earth and M is the mass of the earth find radius of curvature at the point of minimum distance.
These both situation is truely shown in graph (D) A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0 , the value of acceleration due to gravity at the earth's surface, is
An artificial satellite is revolving around a planet of mass M and radius R in a circular orbit of radius r. From Kepler's third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis that T=(k)/(R) sqrt((r^(3))/(g)) where k is dimensionless RV g constant and g is acceleration due to gravity.
