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.
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.
Show Hint
Projected with escape velocity, the speed at distance \(r\) is \(\sqrt{2GM/r}\); integrate \(dt = dr/v\) from \(R_e\) to \(R_e + h\).
Updated On: Nov 4, 2026
- \(\sqrt{\dfrac{2R_e}{g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\)
- \(\dfrac{1}{3}\sqrt{\dfrac{R_e}{2g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\)
- \(\sqrt{\dfrac{R_e}{2g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\)
- \(\dfrac{1}{3}\sqrt{\dfrac{2R_e}{g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Escape speed means total energy is zero, so at distance \(r\) from the centre: \(\tfrac12 v^2 = \dfrac{GM}{r}\), i.e. \(v = \sqrt{\dfrac{2GM}{r}} = R_e\sqrt{\dfrac{2g}{r}}\) (using \(GM = gR_e^2\)).
\(\dfrac{dr}{dt} = R_e\sqrt{2g}\, r^{-1/2}\) so \(r^{1/2}dr = R_e\sqrt{2g}\,dt\).
Integrate from \(R_e\) to \(R_e + h\): \(\tfrac23\left[(R_e + h)^{3/2} - R_e^{3/2}\right] = R_e\sqrt{2g}\; t\).
\(t = \dfrac{2}{3}\dfrac{R_e^{3/2}}{R_e\sqrt{2g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right] = \dfrac23\sqrt{\dfrac{R_e}{2g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\).
Since \(\dfrac23\sqrt{\dfrac{R_e}{2g}} = \dfrac13\sqrt{\dfrac{4R_e}{2g}} = \dfrac13\sqrt{\dfrac{2R_e}{g}}\), \(t = \dfrac13\sqrt{\dfrac{2R_e}{g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\).
Answer: (4) \(\dfrac{1}{3}\sqrt{\dfrac{2R_e}{g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\)
\(\dfrac{dr}{dt} = R_e\sqrt{2g}\, r^{-1/2}\) so \(r^{1/2}dr = R_e\sqrt{2g}\,dt\).
Integrate from \(R_e\) to \(R_e + h\): \(\tfrac23\left[(R_e + h)^{3/2} - R_e^{3/2}\right] = R_e\sqrt{2g}\; t\).
\(t = \dfrac{2}{3}\dfrac{R_e^{3/2}}{R_e\sqrt{2g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right] = \dfrac23\sqrt{\dfrac{R_e}{2g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\).
Since \(\dfrac23\sqrt{\dfrac{R_e}{2g}} = \dfrac13\sqrt{\dfrac{4R_e}{2g}} = \dfrac13\sqrt{\dfrac{2R_e}{g}}\), \(t = \dfrac13\sqrt{\dfrac{2R_e}{g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\).
Answer: (4) \(\dfrac{1}{3}\sqrt{\dfrac{2R_e}{g}}\left[\left(1 + \dfrac{h}{R_e}\right)^{3/2} - 1\right]\)
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Gravitation Questions
- A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution ?
- If \( R \) is the radius of the earth and the acceleration due to gravity on the surface of the earth is \( g = \pi^2 \, \text{m/s}^2 \), then the length of the second's pendulum at a height \( h = 2R \) from the surface of the earth will be:
- Four identical particles of mass \( m \) are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is \[ \left( \frac{2 \sqrt{2} + 1}{32} \right) \frac{G m^2}{L^2}, \] the length of the sides of the square is:
- Assertion: The angular velocity of the moon revolving around the earth is more than that of the earth revolving around the Sun.
Reason: The time taken by the moon to revolve around the earth is less than the time taken by the earth to revolve around the sun. - A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given \( g = 9.8 \, \text{m/s}^2 \))
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions