
The mass of the sun is \[2 \times {10^{30}}kg\] and the mass of the earth is$6 \times {10^{24}}kg$ . if the average distance between the sun and the earth be \[1.5 \times {10^8}km\] , calculate the force of gravitation between them.
Answer
505.2k+ views
Hint: The force of gravitation between two masses is directly proportional to the product of the two masses and inversely proportional to the square of the distance between the two masses. On the basis of the relations, there exists a formula of gravitational force having a constant called Gravitational Constant. Using this formula the force of gravitation can be found if the value of the gravitational constant is known.
Complete step-by-step solution:
According to Newton’s law of universal gravitation, two objects with bigger masses can attract each other by a force called Gravitational force or force of gravitation.
The force of gravitation between two masses is directly proportional to the product of the two masses i.e $F \propto ({M_s} \times {M_e})$ and inversely proportional to the square of the distance between the two masses i.e $F \propto \dfrac{1}{{{d^2}}}$
$F = $ The force of gravitation,
${M_s} = $mass of the sun,
${M_e} = $mass of the earth,
$d = $ the average distance between the sun and the earth.
From this, we get the formula,
$F = G\dfrac{{{M_s} \times {M_e}}}{{{d^2}}}$, $G = $ the gravitational constant = \[6.674 \times {10^{ - 11}}N{\left( {m/kg} \right)^2}\]
Given, \[{M_s} = 2 \times {10^{30}}kg\]
${M_e} = $$6 \times {10^{24}}kg$
$d = $\[1.5 \times {10^8}km = 1.5 \times {10^{11}}m\]
$\therefore F = 6.674 \times {10^{ - 11}}\dfrac{{2 \times {{10}^{30}} \times 6 \times {{10}^{24}}}}{{{{\left( {1.5 \times {{10}^{11}}} \right)}^2}}}$
$ \Rightarrow F = 35.6 \times {10^{21}}$
Hence, the force of gravitation between the sun and the earth is $35.6 \times {10^{21}}N$.
Note: The constant of gravitation is that the proportion constant utilized in the law of motion of Universal Gravitation and is usually denoted by\[G\] . This is often totally different from\[g\] , which denotes the gravitational acceleration. The Law applies to all or any objects with huge or little masses. Two huge objects are thought about as point-like mass, if the space between them is incredibly massive compared to their sizes or if they're spherically symmetric. For these cases, the mass of every object is taken as a point mass set at its center of mass.
Complete step-by-step solution:
According to Newton’s law of universal gravitation, two objects with bigger masses can attract each other by a force called Gravitational force or force of gravitation.
The force of gravitation between two masses is directly proportional to the product of the two masses i.e $F \propto ({M_s} \times {M_e})$ and inversely proportional to the square of the distance between the two masses i.e $F \propto \dfrac{1}{{{d^2}}}$
$F = $ The force of gravitation,
${M_s} = $mass of the sun,
${M_e} = $mass of the earth,
$d = $ the average distance between the sun and the earth.
From this, we get the formula,
$F = G\dfrac{{{M_s} \times {M_e}}}{{{d^2}}}$, $G = $ the gravitational constant = \[6.674 \times {10^{ - 11}}N{\left( {m/kg} \right)^2}\]
Given, \[{M_s} = 2 \times {10^{30}}kg\]
${M_e} = $$6 \times {10^{24}}kg$
$d = $\[1.5 \times {10^8}km = 1.5 \times {10^{11}}m\]
$\therefore F = 6.674 \times {10^{ - 11}}\dfrac{{2 \times {{10}^{30}} \times 6 \times {{10}^{24}}}}{{{{\left( {1.5 \times {{10}^{11}}} \right)}^2}}}$
$ \Rightarrow F = 35.6 \times {10^{21}}$
Hence, the force of gravitation between the sun and the earth is $35.6 \times {10^{21}}N$.
Note: The constant of gravitation is that the proportion constant utilized in the law of motion of Universal Gravitation and is usually denoted by\[G\] . This is often totally different from\[g\] , which denotes the gravitational acceleration. The Law applies to all or any objects with huge or little masses. Two huge objects are thought about as point-like mass, if the space between them is incredibly massive compared to their sizes or if they're spherically symmetric. For these cases, the mass of every object is taken as a point mass set at its center of mass.
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