Derive an expression for binding energy of a body at rest on the Earth's surface.

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Accepted Answer

A body of mass m at rest on the Earth's surface has kinetic energy zero and potential energy $U(R)=-\dfrac{GMm}{R}$ (using the standard $U=-GMm/r$ formula at $r=R$ ), so its total mechanical energy is $TE=0+\left(-\dfrac{GMm}{R}\right)=-\dfrac{GMm}{R}.$ This total energy is negative, meaning the body is gravitationally bound to the Earth. To free it -- to just barely remove it to infinity, where $U(\infty)=0$ and, for the MINIMUM removal energy, the final kinetic energy is also taken as zero -- exactly enough energy must be supplied to raise the total energy from $-GMm/R$ up to zero. This required energy is the binding energy of the body: $BE=0-TE=0-\left(-\dfrac{GMm}{R}\right)=\dfrac{GMm}{R}.$ Equivalently, this is exactly the kinetic energy the body would need at launch to just …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the escape velocity of earth (approximately)?$
(a) 10 m/s
(b) 100 m/s
(c) 1000 m/s
(d) 11.2 km/s
›Reveal solutionSolution

Earth's escape velocity, the minimum speed needed to escape Earth's gravitational pull without further propulsion, is approximately 11.2 km/s.

Escape velocity is derived by equating the kinetic energy needed to the gravitational potential energy binding the object to Earth:

(1/2) m v_e^2 = G M m / R

v_e = sqrt(2GM/R)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Case study: We have learnt that the earth attracts objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall. Is there any change in the velocity of falling objects? While falling, there is no change in the direction of motion of the objects. But due to earth's attraction, there will be a change in the magnitude of velocity. Any change in the velocity involves acceleration; whenever an object falls towards the earth, an acceleration is involved. Escape velocity of a body does not depend on:$
(a) Mass of body
(b) Mass of Earth
(c) Radius of Earth
(d) Universal Gravitational Constant
›Reveal solutionSolution

Setting the escaping body's total energy to zero, $\tfrac12mv_e^2=\dfrac{GMm}{R}$ , the mass $m$ of the body cancels — escape velocity is independent of it.

Escape velocity is the minimum launch speed for a body to just escape a planet's gravitational pull, reaching infinity with zero residual kinetic energy. Setting the body's total mechanical energy at launch equal to zero (the energy required to 'just escape'):

$\frac12 m v_e^2 - \frac{GMm}{R} = 0 \quad\Rightarrow\quad \frac12 m v_e^2 = \frac{GMm}{R}$

The mass $m$ of the escaping body appears on both sides and cancels completely:

$v_e = \sqrt{\frac{2GM}{R}}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Moon has no atmosphere because -$
(a) it is close to earth
(b) it revolves around earth
(c) it has very low temperature
(d) it has low value of escape velocity
›Reveal solutionSolution

A low escape velocity lets gas molecules easily reach the speed needed to leave a body's gravity permanently, which is why the Moon cannot hold an atmosphere.

Escape velocity is v_e = sqrt(2GM/R). Since the Moon's mass M and radius R are both much smaller than Earth's, its escape velocity works out to only about 2.38 km/s, compared with Earth's 11.2 km/s.

Gas molecules in any atmosphere have a range of thermal speeds (root-mean-square speed depends on temperature and molar mass). On the Moon, a much larger fraction of gas molecules exceed the (low) escape velocity than on Earth, so over geological time essentially all atmospheric gas molecules have leaked away into space. This is a general rule: any body whose escape velocity is comparable to or less than the thermal speeds of common gas molecules cannot retain an atmosphere.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write true or false: The value of escape velocity at the poles of the earth is less than the escape velocity from the equator.$
›Reveal solutionSolution

The statement is FALSE: escape velocity at the poles is slightly greater, not less, than at the equator.

Escape velocity from Earth's surface is:

v_e = sqrt(2GM / R)

where M is Earth's total mass (essentially the same regardless of where you stand on the surface) and R is the distance from the centre of the Earth to that point on the surface.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If Ve is escape velocity and Vo is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by$
(a) Vo = sqrt(2) Ve
(b) Vo = Ve
(c) Ve = Vo / 2
(d) Ve = sqrt(2) Vo
›Reveal solutionSolution

Ve = sqrt(2GM/R) and Vo = sqrt(GM/R) for a near-surface orbit, so their ratio is exactly Ve/Vo = sqrt(2), i.e. Ve = sqrt(2) Vo.

Orbital velocity (for a satellite in a circular orbit just above Earth's surface, radius ~ R):

Vo = sqrt(GM/R)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The escape velocity of a body of mass 'm' is given by$
(a) $\sqrt{GM/R}$
(b) $\sqrt{2GM/R}$
(c) $\sqrt{2GMm/R}$
(d) $\sqrt{(2GM+R)/R}$
›Reveal solutionSolution

Escape velocity is the minimum speed needed so that a body's total mechanical energy at the surface is $\ge 0$ , i.e. it can just reach infinity with zero kinetic energy.

At the surface, $KE=\dfrac12 mv_e^2$ and gravitational PE $=-\dfrac{GMm}{R}$ .

For the body to just escape (total energy at infinity $=0$ ):

$\dfrac12 mv_e^2 - \dfrac{GMm}{R} = 0$

…

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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: The escape velocity from moon surface is nearly ......... km s^-1.$
›Reveal solutionSolution

The escape velocity from the Moon's surface is about 2.4 km/s, much smaller than Earth's 11.2 km/s.

Escape velocity is given by v_e = sqrt(2GM/R), where M and R are the mass and radius of the body. The Moon has a much smaller mass (about 1/81 of Earth's) and smaller radius (about 0.27 times Earth's) than Earth, so GM/R for the Moon is much lower than for Earth. Using the Moon's actual mass and radius, v_e works out to approximately 2.38 km/s, which is why it is easier for gas molecules (and an …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The escape velocity of a body released from the surface of the earth is: (where RE is the radius of the earth and g is the acceleration due to gravity at the earth's surface)$
(a) √(gRE)
(b) √(2gRE)
(c) 2√(gRE)
(d) √(gRE/2)
›Reveal solutionSolution

The escape velocity from the earth's surface is ve = √(2 g RE), obtained by equating the launch kinetic energy to the gravitational binding energy.

For a body of mass m to just escape the earth's gravitational pull, its initial kinetic energy at the surface must equal the magnitude of its gravitational potential energy there (so that total mechanical energy is zero, allowing it to reach infinity with zero residual speed):

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Match the column: Escape velocity ve — match with the correct expression.$
(a) sqrt(2gR)
(b) sqrt(T/m)
(c) GMm/r^2
(d) I*omega
(e) 2pisqrt(l/g)
(f) sqrt(gR)
(g) m*R^2
›Reveal solutionSolution

Escape velocity from a planet's surface is ve = sqrt(2gR), matching option (a) -- exactly sqrt(2) times the orbital speed sqrt(gR).

Escape velocity is the minimum launch speed needed for an object to escape a planet's gravity completely, reaching infinity with zero final kinetic energy. Equating initial kinetic energy to the (magnitude of) gravitational potential energy at the surface:

(1/2)mve^2 = GMm/R

…

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CBSE 2022Set ANN1 mark
$Q. Write the value of escape velocity of an object from the surface of moon.$
›Reveal solutionSolution

The escape velocity from the Moon's surface is about 2.4 km/s, much smaller than Earth's 11.2 km/s because the Moon has far smaller mass and radius.

Escape velocity is v_e = √(2GM/R), where M and R are the mass and radius of the body. The Moon's mass (~7.3 × 10²² kg) and radius (~1737 km) are both much smaller than Earth's, and the mass term dominates the ratio M/R, so v_e works out far smaller for the Moon than for Earth. This low escape velocity is also the physical reason the Moon effective …

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CBSE 2022Set TERM11 markMCQ
$Q. The escape velocity from the surface of the earth is (Re is radius of earth)$
(1) sqrt(g Re)
(2) sqrt(2 g Re)
(3) sqrt(3 g Re)
(4) sqrt(4 g Re)
›Reveal solutionSolution

Escape velocity is found by requiring the object's kinetic energy to exactly equal the magnitude of its gravitational potential energy at the surface, (1/2)mv_e^2 = GMm/Re; using g = GM/Re^2 simplifies this to v_e = sqrt(2 g Re).

For an object of mass m to just escape Earth's gravitational field from the surface (reaching infinity with zero residual speed), its initial kinetic energy must equal the magnitude of the gravitational potential energy binding it:

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Escape velocity of a body from the Earth depends on:$
(a) Mass of the body
(b) Direction of projection
(c) Position of the projection point
(d) Height of the projection point from the earth
›Reveal solutionSolution

Escape velocity ve = sqrt(2GM/r) depends on the mass M of the Earth and the distance r of the projection point from the Earth's centre — so it changes with the height of launch, but not with the mass or direction of the projected body.

Derivation: for a body of mass m projected with escape velocity from a distance r from Earth's centre, the minimum kinetic energy needed equals the magnitude of gravitational potential energy at that point: (1/2)mve^2 = GMm/r, giving ve = sqrt(2GM/r). Since r = R + h (R = Earth's radius, h = height above surface), a body launched from a greater height h needs a smaller escape velocity. Notice m (mass of the body) cancels out, and the formula has no direction dependence, s …

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Q.Derive an expression for binding energy of a body at rest on the Earth's surface.

Concept understanding — Escape Velocity

Q.Derive an expression for binding energy of a body at rest on the Earth's surface.

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