10m/s2
9.8m/s2
20m/s2
2.5m/s2
5m/s2
Given:
Step 1: Relate Escape Speed to Gravity
Escape speed is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant and \( M \) is the planet's mass.
Step 2: Express Surface Gravity
Acceleration due to gravity at the surface (\( g \)) is:
\[ g = \frac{GM}{R^2} \]
Step 3: Solve for \( g \) Using Escape Speed
Square the escape speed equation:
\[ v_e^2 = \frac{2GM}{R} \implies GM = \frac{v_e^2 R}{2} \]
Substitute \( GM \) into the gravity formula:
\[ g = \frac{v_e^2 R}{2 R^2} = \frac{v_e^2}{2R} \]
Plug in the given values:
\[ g = \frac{(10,000)^2}{2 \times 10^7} = \frac{10^8}{2 \times 10^7} = 5 \, \text{m/s}^2 \]
Conclusion:
The acceleration due to gravity at the planet's surface is 5 m/s\(^2\).
Answer: \(\boxed{E}\)
1. Recall the formula for escape speed:
The escape speed (ve) of a planet is given by:
\[v_e = \sqrt{\frac{2GM}{R}}\]
where:
2. Relate escape speed to acceleration due to gravity:
The acceleration due to gravity (g) at the surface of a planet is given by:
\[g = \frac{GM}{R^2}\]
3. Combine the equations and solve for g:
From the escape speed equation, we can write:
\[v_e^2 = \frac{2GM}{R}\]
Now, multiply both sides by R:
\[v_e^2 R = 2GM\]
Divide both sides by 2R²:
\[\frac{v_e^2}{2R} = \frac{GM}{R^2} = g\]
4. Plug in the given values:
We are given \(v_e = 10 \, km/s = 10^4 \, m/s\) and \(R = 10,000 \, km = 10^7 \, m\). Substituting these values into the equation for g:
\[g = \frac{(10^4 \, m/s)^2}{2 \times 10^7 \, m} = \frac{10^8 \, m^2/s^2}{2 \times 10^7 \, m} = 5 \, m/s^2\]
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: 
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:
Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec.
The formula for escape speed is given below:
ve = (2GM / r)1/2
where ,
ve = Escape Velocity
G = Universal Gravitational Constant
M = Mass of the body to be escaped from
r = Distance from the centre of the mass