Find the gravitational force between two ston | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: Mass of each stone, $m_1 = m_2 = 2 \,kg$.Separation distance, $r = 1 \,m$.Universal gravitational constant, $G = 6.67 	imes 10^{-11} \,N \

Vedclass · Accepted Answer

$2.67 	imes 10^{-10} \,N$

Vedclass · Answer

(D) Given: Mass of each stone, $m_1 = m_2 = 2 \,kg$.Separation distance, $r = 1 \,m$.Universal gravitational constant, $G = 6.67 	imes 10^{-11} \,N \cdot m^2/kg^2$.According to Newton's law of gravitation, the force $F$ is given by $F = \frac{G m_1 m_2}{r^2}$.Substituting the values: $F = \frac{6.67 	imes 10^{-11} 	imes 2 	imes 2}{1^2}$.$F = 6.67 	imes 10^{-11} 	imes 4$.$F = 26.68 	imes 10^{-11} \,N = 2.668 	imes 10^{-10} \,N$.Rounding to the nearest value, $F \approx 2.67 	imes 10^{-10} \,N$.

Find the gravitational force between two stones, each of mass $2 \,kg$ and separated by a distance $1 \,m$ in vacuum.

Similar Questions

If the masses of two point objects are tripled and the distance between them is doubled,then the gravitational force of attraction between them will increase by .......... $\%$

$A$ mass $M$ is divided into two parts $xM$ and $(1 - x)M$. For a given separation,the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is

$A$ large number of identical point masses $m$ are placed along the $x$-axis at $x = 0, 1, 2, 4, 8, \dots$ units. The magnitude of the total gravitational force on the mass at the origin $(x = 0)$ will be:

The force of mutual attraction between any two objects by virtue of their masses is

Gravitational forces operate among which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module