
What is the weight of a 70 kg body on the surface of a planet whose mass is ${{\dfrac{1}{7}}^{th}}$ that of the earth and radius is half of the earth?
A. 20 kg
B. 40 kg
C. 70 kg
D. 140 kg
Answer
125.7k+ views
Hint: Weight of a body on any planet is given by mg, where g is the gravitational acceleration on the planet. So, to find the weight of this body on another planet we need to find the gravitational acceleration g’ on the surface of that planet arising due to the gravitational attraction. Which depends on the mass and the radius of the planet.
Complete step-by-step answer:
Gravitational attraction force, which is essentially the weight measured on a planet, exerted by the planet on a body of mass m is given as:
$\overset{\to }{\mathop{F}}\,=\dfrac{GMm}{{{R}^{2}}}$
Where,
M is the mass of the planet
m is the mass of body
R is the radius of the planet
G is the gravitational constant
Since we know that weight is also represented mathematically as mg.
Let us take the mass of earth as M and radius as R. Therefore, the acceleration experienced by the object of mass m on earth will be g:
$g=\dfrac{GM}{{{R}^{2}}}$
According to the question:
Mass of that planet M’ $=\dfrac{M}{7}$ where M is the mass of earth
Radius of that planet R’ $=\dfrac{R}{2}$ where R is the Radius of earth
Now we can put these values in the above given equation as;
$\begin{align}
& g'=\dfrac{GM'}{{{R}^{'2}}} \\
& \Rightarrow g'=\dfrac{4GM}{7{{R}^{2}}} \\
& \Rightarrow g'=\dfrac{4}{7}g \\
\end{align}$
As we know that mg = 70kg is the weight on earth. Multiplying m on both sides:
$\begin{align}
& mg'=\dfrac{4}{7}mg \\
& \Rightarrow mg'=\dfrac{4}{7}70kg=40kg \\
\end{align}$
So, the weight of a 70 kg body on the given planet will be 40kg. Option B.
Note: Gravitational force is a conservative force, which means that work done by this force doesn’t depend on the path followed. Gravitational force’s effect decreases with increase in distance, and increases with increase in mass.
Complete step-by-step answer:
Gravitational attraction force, which is essentially the weight measured on a planet, exerted by the planet on a body of mass m is given as:
$\overset{\to }{\mathop{F}}\,=\dfrac{GMm}{{{R}^{2}}}$
Where,
M is the mass of the planet
m is the mass of body
R is the radius of the planet
G is the gravitational constant
Since we know that weight is also represented mathematically as mg.
Let us take the mass of earth as M and radius as R. Therefore, the acceleration experienced by the object of mass m on earth will be g:
$g=\dfrac{GM}{{{R}^{2}}}$
According to the question:
Mass of that planet M’ $=\dfrac{M}{7}$ where M is the mass of earth
Radius of that planet R’ $=\dfrac{R}{2}$ where R is the Radius of earth
Now we can put these values in the above given equation as;
$\begin{align}
& g'=\dfrac{GM'}{{{R}^{'2}}} \\
& \Rightarrow g'=\dfrac{4GM}{7{{R}^{2}}} \\
& \Rightarrow g'=\dfrac{4}{7}g \\
\end{align}$
As we know that mg = 70kg is the weight on earth. Multiplying m on both sides:
$\begin{align}
& mg'=\dfrac{4}{7}mg \\
& \Rightarrow mg'=\dfrac{4}{7}70kg=40kg \\
\end{align}$
So, the weight of a 70 kg body on the given planet will be 40kg. Option B.
Note: Gravitational force is a conservative force, which means that work done by this force doesn’t depend on the path followed. Gravitational force’s effect decreases with increase in distance, and increases with increase in mass.
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