चंद्राचे वस्तुमान व त्रिज्या अनुक्रमे \( 7.34 \times 10^{22} \, kg \) व \( 1.74 \times 10^{6} \, m \) आहेत. चंद्रावरील गुरुत्व त्वरण काढा.
\[ (G = 6.67 \times 10^{-11} \, Nm^{2}/kg^{2}) \]
Step 1: सूत्र लिहा.
चंद्रावरील गुरुत्व त्वरण काढण्यासाठी सूत्र आहे —
\[
g = \frac{G M}{R^{2}}
\]
इथे,
\( G = 6.67 \times 10^{-11} \, Nm^{2}/kg^{2} \),
\( M = 7.34 \times 10^{22} \, kg \),
\( R = 1.74 \times 10^{6} \, m \).
Step 2: मूल्ये घाला.
\[
g = \frac{6.67 \times 10^{-11} \times 7.34 \times 10^{22}}{(1.74 \times 10^{6})^{2}}
\]
\[
g = \frac{4.89 \times 10^{12}}{3.02 \times 10^{12}} = 1.62 \, m/s^{2}
\]
Step 3: निष्कर्ष.
चंद्रावरील गुरुत्व त्वरणाचे मूल्य 1.62 m/s² आहे, जे पृथ्वीवरील गुरुत्व त्वरणापेक्षा सुमारे 1/6 पट कमी आहे.
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:
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: 