Mock Test 1 — Gravitation

The acceleration due to gravity at a depth $d$ from Earth's surface is half the surface value. The value of $d$:

The orbital velocity of an artificial satellite very close to Earth's surface is:

A particle of mass $m$ is taken from Earth's surface to a height equal to $R$. The work done:

The escape velocity of a body from a planet depends on:

Two planets have radii $r_1, r_2$ and densities $\rho_1, \rho_2$. The ratio of $g$ on their surfaces:

The gravitational PE of a body at Earth's surface is $-mgR$. Its PE at height $h = R$ above surface:

A satellite of mass $m$ revolves around Earth at altitude $h$. Its angular momentum:

If $g$ on Earth is $9.8$ m/s² and Earth's radius is $6400$ km, then escape velocity:

The time period of a satellite in a circular orbit of radius $r$ around a planet of mass $M$:

A satellite is moving in a circular orbit with kinetic energy $K$. Its potential energy is:

The ratio of acceleration due to gravity at depth $h$ to height $h$ above Earth's surface (both small, $h \ll R$) is $\dfrac{x-h/R}{1-2h/R}$ where the value of $x$ is:

A satellite is at height $3R$ above Earth's surface ($R$ = radius). The orbital velocity is $v_0/\sqrt{n}$ where $v_0$ is the orbital velocity near Earth's surface. The value of $n$ is:

Escape velocity from a planet of radius $R$ is $v$. From a planet of same density but radius $2R$, escape velocity in terms of $v$ (as integer multiple) is:

Assertion (A): A satellite in orbit experiences no gravitational force. Reason (R): In orbit, gravity is balanced by centrifugal force in the satellite's frame.

Assertion (A): The total energy of a satellite in circular orbit is negative. Reason (R): The kinetic energy is half the magnitude of the potential energy.