A space ship of mass 2 × 104 kg is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be (if g = 10 m/s2 and radius of earth = 6400 km)
- 11.2 (√2 -1) km/s
- 7.9 (√2 - 1) km/s
- 8(√2 -1) km/s
- 7.4(√2 - 1) km/s
The Correct Option is C
Solution and Explanation
Step 1: Formula for \( \Delta V \)
The change in velocity (\( \Delta V \)) is given by:
\[ \Delta V = \frac{GM}{R} (\sqrt{2} - 1), \]
where:
- \( G \): Gravitational constant
- \( M \): Mass of the central body (e.g., Earth)
- \( R \): Radius of the orbit
Step 2: Simplify the Expression
The term \( \frac{GM}{R} \) can be rewritten using the acceleration due to gravity at the surface of the central body (\( g \)):
\[ g = \frac{GM}{R^2}. \]
Substitute \( gR \) for \( \frac{GM}{R} \):
\[ \Delta V = gR (\sqrt{2} - 1). \]
Step 3: Substitute Numerical Values
We are given \( gR = 8000 \, \text{m/s} \) or \( 8 \, \text{km/s} \). Substitute this into the equation:
\[ \Delta V = 8000 (\sqrt{2} - 1) \, \text{m/s}. \]
Convert to km/s:
\[ \Delta V = 8 (\sqrt{2} - 1) \, \text{km/s}. \]
Final Answer:
\( \Delta V = 8 (\sqrt{2} - 1) \, \text{km/s} \).
Top Questions on Gravitation
- Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ______ seconds.
- JEE Main - 2025
- Physics
- Gravitation
- A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.
- JEE Main - 2025
- Physics
- Gravitation
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
- JEE Main - 2025
- Physics
- Gravitation
- Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.
Reason (R) : For a central force field the angular momentum is a constant. In the light of the above statements, choose the most appropriate answer from the options given below :- JEE Main - 2025
- Physics
- Gravitation
- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize} \item[(A)] A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. \item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet. \end{itemize} In light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2025
- Physics
- Gravitation
Questions Asked in JEE Main exam
- Two water drops each of radius \( r \) coalesce to form a bigger drop. If \( T \) is the surface tension, the surface energy released in this process is:
- JEE Main - 2025
- Surface Tension
- An AC current is represented as: $ i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right) \text{ Amp} $ The RMS value of the current is:
- JEE Main - 2025
- AC Circuits
- A particle of mass \( m \) and charge \( q \) is fastened to one end \( A \) of a massless string having equilibrium length \( l \), whose other end is fixed at point \( O \). The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field is switched on along the direction as shown in the figure, then the speed of the particle when it crosses the x-axis is:
- JEE Main - 2025
- Electric charges and fields
- An electric bulb rated as 100 W-220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :
- JEE Main - 2025
- AC Circuits
- If \[ \sum_{r=1}^{13} \frac{1}{\sin \frac{\pi}{4} + (r-1) \frac{\pi}{6}} \sin \frac{\pi}{4} + \frac{\pi}{6} = a \sqrt{3} + b, \quad a, b \in \mathbb{Z}, { then } a^2 + b^2 { is equal to:} \]
- JEE Main - 2025
- Trigonometry