A body is projected vertically upwards from the surface of Earth with a velocity sufficient enough t

Q: Why is the total energy zero?

When a body is projected with exactly the escape velocity, it reaches infinity with zero kinetic energy. Since potential energy at infinity is also zero, the total mechanical energy is zero. By conservation of energy, it must be zero everywhere.

Q: Why can't we use the standard equations of motion like $s = ut + \frac{1}{2}at^2$?

Those equations are only valid for constant acceleration. Here, the acceleration due to gravity $g$ decreases as the body moves further away from the Earth, so we must use calculus.

Q: Why did we use $r$ instead of $h$ for the integration?

The gravitational force depends on the distance from the center of the Earth, $r$. Using $r$ makes the integration straightforward. We can always substitute $r = R_e + h$ at the end.

Q: What if the body was projected with a velocity greater than the escape velocity?

In that case, the total energy would be positive. The body would reach infinity with some residual kinetic energy, and the integration would involve a more complex function.

Q: How did we replace $GM$ with $g$?

The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R_e^2}$. Rearranging this gives $GM = gR_e^2$, which is a handy substitution to match the options.

Question

A body is projected vertically upwards from the surface of Earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is ......... s.

Options:

(A) $\sqrt{\frac{R_e}{2g}} \left[ \left(1 + \frac{h}{R_e}\right)^{3/2} - 1 \right]$

(B) $\sqrt{\frac{2R_e}{g}} \left[ \left(1 + \frac{h}{R_e}\right)^{3/2} - 1 \right]$

(C) $\frac{1}{3}\sqrt{\frac{R_e}{2g}} \left[ \left(1 + \frac{h}{R_e}\right)^{3/2} - 1 \right]$

(D) $\frac{1}{3}\sqrt{\frac{2R_e}{g}} \left[ \left(1 + \frac{h}{R_e}\right)^{3/2} - 1 \right]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing on the surface of the Earth, and you project a body vertically upwards. The problem states that the velocity is "sufficient enough to carry it to infinity." This is a crucial piece of information! It tells us that the body has been projected with exactly the **escape velocity**.

Because it just barely reaches infinity, its velocity at infinity will be zero. Since the potential energy at infinity is also zero, the **total mechanical energy** of the body is zero at the end of its journey. By the law of conservation of energy, the total energy must be zero at *every* point during its flight.

### The Master Equation
Let's apply this conservation of energy at an arbitrary distance $r$ from the center of the Earth. The sum of kinetic energy and potential energy must equal zero:
$$ \frac{1}{2}mv^2 - \frac{GMm}{r} = 0 $$

From this, we can easily find the velocity $v$ as a function of the distance $r$:
$$ v = \sqrt{\frac{2GM}{r}} $$

Now, velocity is simply the rate of change of position, so we can write $v = \frac{dr}{dt}$. Substituting this into our equation gives us a differential equation:
$$ \frac{dr}{dt} = \sqrt{\frac{2GM}{r}} $$

### Setting Up the Integral
To find the time $t$ it takes to reach a height $h$ above the surface, we need to separate the variables and integrate. Let's move all the $r$ terms to one side and the $dt$ to the other:
$$ dt = \frac{1}{\sqrt{2GM}} r^{1/2} dr $$

Now, we integrate both sides. At time $t = 0$, the body is at the surface of the Earth, so $r = R_e$. At time $t$, the body is at a height $h$ above the surface, so its distance from the center is $r = R_e + h$.
$$ \int_0^t dt = \frac{1}{\sqrt{2GM}} \int_{R_e}^{R_e+h} r^{1/2} dr $$

### Final Calculation
The integration of $r^{1/2}$ is straightforward—it becomes $\frac{r^{3/2}}{3/2}$. Applying the limits, we get:
$$ t = \frac{1}{\sqrt{2GM}} \left[ \frac{2}{3} r^{3/2} \right]_{R_e}^{R_e+h} $$
$$ t = \frac{2}{3\sqrt{2GM}} \left[ (R_e+h)^{3/2} - R_e^{3/2} \right] $$

We are almost there! The options are given in terms of the acceleration due to gravity at the surface, $g$. We know the standard relation $g = \frac{GM}{R_e^2}$, which means we can substitute $GM = gR_e^2$ into our equation:
$$ t = \frac{2}{3\sqrt{2gR_e^2}} \left[ (R_e+h)^{3/2} - R_e^{3/2} \right] $$

Let's pull $R_e$ out of the square root in the denominator, and factor out $R_e^{3/2}$ from the numerator:
$$ t = \frac{2 R_e^{3/2}}{3 R_e \sqrt{2g}} \left[ \left(1 + \frac{h}{R_e}\right)^{3/2} - 1 \right] $$

Simplifying the powers of $R_e$, we arrive at our final, elegant expression:
$$ t = \frac{1}{3} \sqrt{\frac{2R_e}{g}} \left[ \left(1 + \frac{h}{R_e}\right)^{3/2} - 1 \right] $$

This perfectly matches option (d). Take a moment to appreciate how conservation of energy and a simple integration elegantly solved what seemed like a complex kinematics problem!

Animated Solution for Physics - Gravitation: A body is projected vertically upwards from the surface of Earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is ......... s.

Select Answer:

Visualized Solution

Initial Setup

Energy Conservation

Velocity as a function of $r$

Setting up the Integral

Evaluating the Integral

Substituting $g$

Final Simplification

Conclusion

The Sigma Insight: Escape Speed and Motion of Satellites

Analyzing the Setup Imagine you are standing on the surface of the Earth, and you project a body vertically upwards

The Master Equation Let's apply this conservation of energy at an arbitrary distance $r$ from the center of the Earth

Setting Up the Integral To find the time $t$ it takes to reach a height $h$ above the surface, we need to separate the variables and integrate

Final Calculation The integration of $r^{1/2}$ is straightforward—it becomes $\frac{r ^{3/2}}{3/2}$

Similar Questions

The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$ ) to infinity is

The escape velocity for a body projected vertically upwards from the surface of earth is $11 km/s$ . If the body is projected at an angle of $4 5^{\circ}$ with the vertical, then the escape velocity will be

A satellite is revolving in a circular orbit at a height $h$ from the Earth's surface (radius of Earth $R, h ≪ R$ ). The minimum increase in its orbital velocity required, so that the satellite could escape from the Earth's gravitational field, is close to (Neglect the effect of atmosphere)

A satellite is moving with a constant speed $v$ in circular orbit around the earth. An object of mass ' $m$ ' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is

The masses and radii of the Earth and Moon are $(m_{1}, R_{1})$ and $(m_{2}, R_{2})$ , respectively. Their centres are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the middle of these two masses.

The radius in kilometre to which the present radius of Earth ( $R = 6400 km$ ) to be compressed so that the escape velocity is increased 10 times 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 $(\frac{v}{2})$ the instantaneous velocity $v$ of $A$ . The collision is completely inelastic. Then, the combined body

Planet A has mass $M$ and radius $R$ . Planet B has half the mass and half the radius of planet A. If the escape velocities from the planets A and B are $v_{A}$ and $v_{B}$ respectively, then $\frac{v _{A}}{v _{B}} = \frac{n}{4}$ . The value of $n$ is