Satellite Energy & Weightlessness

The kinetic energy of a revolving satellite (mass $m$) at a height equal to thrice the radius of the earth ($R$) is ($g$ = acceleration due to gravity)

The minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $3R$ is $\{G$ $=$ gravitational constant$\}$

Initially a satellite of 100 kg is in a circular orbit of radius $1.5R_E$. This satellite can be moved to a circular orbit of radius $3R_E$ by supplying $\alpha \times 10^{6}\text{ J}$ of energy. The value of $\alpha$ is ______. (Take Radius of Earth $R_E = 6 \times 10^{6}\text{ m}$ and $g = 10\text{ m/s}^{2}$)

The total energy of a circularly orbiting satellite is

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is____________ $\times 10^{10}\text{ J}$. (Mass of earth $= 6 \times 10^{24}\text{ kg}$, Radius of earth $= 6.4 \times 10^{6}\text{ m}$, Gravitational constant $= 6.67 \times 10^{-11}\text{ N m}^{2}\text{ kg}^{-2}$)

For a satellite moving in an orbit around the earth at height ' $h$ ' the ratio of kinetic energy to potential energy is

Two satellites A and B having ratio of masses $3 : 1$ are revolving in circular orbits of radii ' $r$ ' and ' 4 r '. The ratio of total energy of satellites A to that of B is

Match List I with List II:

(Where $a =$ radius of planet orbit, $r =$ radius of planet, $M =$ mass of Sun, $m =$ mass of planet)

Choose the correct answer from the options given below:

The minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$ is

The gravitational potential energy required to raise a satellite of mass ' $m$ ' to height ' $h$ ' above the earth's surface is ' $E_1$ '. Let the energy required to put this satellite into the orbit at the same height be ' $E_2$ '. If $M$ and $R$ are the mass and radius of the earth respectively then $E_1 : E_2$ is

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$ from the surface of the earth is:

A satellite of $10^{3}\text{ kg}$ mass is revolving in circular orbit of radius $2R$. If $\frac{10^{4}R}{6}\text{ J}$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius

(use $g = 10\text{ m/s}^{2}, R =$ radius of earth)

An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5\text{ km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x\frac{GM_e m}{21R_e}$. The value of $x$ is (take $R_e = 6370\text{ km})$:

For a satellite orbiting around the earth in a circular orbit, the ratio of potential energy to kinetic energy at same height is

Given below are two statements:

Statement I: If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $\frac{E}{2}$.

Statement II: The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $E$.

In the light of the above statements, choose the most appropriate answer from the options given below

A seconds pendulum is placed in a space laboratory orbiting round the earth at a height '$3R$' from the earth's surface. The time period of the pendulum will be ( $R =$ radius of earth)

The ratio of energy required to raise a satellite to a height '$h$' above the earth's surface to that required to put it into the orbit at the same height is ($R =$ radius of earth)

Two satellites $A$ and $B$, having masses in the ratio $4 : 3$, are revolving in circular orbits of radii $3r$ and $4r$ respectively around the earth. The ratio of total mechanical energy of $A$ to $B$ is:

Two satellites of same mass are launched in circular orbits at heights '$R$' and '$2R$' above the surface of the earth. The ratio of their kinetic energies is ($R =$ radius of the earth)

The ratio of energy required to raise a satellite of mass '$m$' to height '$h$' above the earth's surface to that required to put it into the orbit at same height is [ $R =$ radius of earth]

The ratio of energy required to raise a satellite of mass $m$ to a height $h$ above the earth's surface of that required to put it into the orbit at same height is [$R =$ radius of the earth]

A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth’s radius Re. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become $\sqrt{\frac{3}{2}}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is:

A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass $\frac{m}{2}$ collides with A with a velocity which is half $\left(\frac{\vec{v}}{2}\right)$ the instantaneous velocity$\vec{v}$of A. The collision is completely inelastic. Then, the combined body:

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass $\frac{m}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth):

The kinetic energy of a revolving satellite (mass $m$) at a height equal to thrice the radius of the earth $(R)$ is

Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB, is;

What is the minimum energy required to launch a satellite of mass ' $m$ ' from the surface of the earth of mass ' $M$ ' and radius ' $R$ ' at an altitude $2R$?

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $\times$ 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is

A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be:

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2},$ and the other mass, in a circular orbit of radius $\frac{3R}{2}$. The difference between the final and initial total energies is:

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 g0, the value of acceleration due to gravity at the earth's surface, is

An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth’s surface. The radius of the earth is R, while its mass is M. The gravitational pull FG on the astronaut is:

What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$?

The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2(R2 > R1) is

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$ is

Gravitational force is required for

The satellite of mass $m$ is orbiting around the earth in a circular orbit with a velocity $v$. What will be its total energy?