Answer
Verified
357.6k+ views
Hint: The potential energy associated with gravitational force, such as rising items against the Earth's gravity, is known as gravitational energy. The gravitational potential energy of a system of respective masses at a certain distance, using gravitational constant $G$ is $U = - \dfrac{{GMm}}{r}$ .
Complete answer:
Let us consider;
Radius of the earth is given by $R$
Mass of the earth is given by $M$
Mass of the given body will be $m$ and
Acceleration due to gravity on earth is given by $g$
When a body is on the surface of the earth, we can write the force of attraction of the earth on it.
${U_1} = - \dfrac{{GMm}}{R}......\left( 1 \right)$
Where, $G = $ Gravitational constant.
When the body is at a height of \[h\] from the earth's surface, the PE of the system is given by
${U_2} = - \dfrac{{GmM}}{{R + h}}$
The potential energy of the object at height $h = R$ from the surface of the earth.
${U_2} = - \dfrac{{GmM}}{{R + R}} = - \dfrac{{GmM}}{{2R}}......\left( 2 \right)$
As a result of the displacement of the body of mass m from the surface to a height equal to the radius \[\left( R \right)\] of the earth, the rise in PE is given by-
$
\Delta U = {U_2} - {U_1} \\
\Delta U = - \dfrac{{GMm}}{{R + R}} + \dfrac{{GMm}}{R} \\
\Delta U = - \dfrac{{GMm}}{{2R}} + \dfrac{{GMm}}{R} \\
\Delta U = \dfrac{1}{2}\dfrac{{GMm}}{R} \\
$
But we know that $GM = g{R^2}$
Hence, $\Delta U = \dfrac{1}{2}\dfrac{{g{R^2}m}}{R}$
Now, one $R$ will be cancelled out so the equation will become
\[
\Delta U = \dfrac{1}{2}gRm \\
\therefore \Delta U = \dfrac{1}{2}mgR \\
\]
Therefore the increase in P.E of a body of mass \['m'\] up to a distance equal to the radius of earth from the earth surface will be $\dfrac{1}{2}mgR$.
Note: It should be noted that the formula for gravitational acceleration is \[g = \dfrac{{GM}}{{{r^2}}}\] . It is determined by the earth's mass and radius. This enables us to comprehend the following:-
1. Gravity accelerates all bodies at the same rate, regardless of their mass.
2. It's worth on Earth is determined by the mass of the planet, not the mass of the object.
Complete answer:
Let us consider;
Radius of the earth is given by $R$
Mass of the earth is given by $M$
Mass of the given body will be $m$ and
Acceleration due to gravity on earth is given by $g$
When a body is on the surface of the earth, we can write the force of attraction of the earth on it.
${U_1} = - \dfrac{{GMm}}{R}......\left( 1 \right)$
Where, $G = $ Gravitational constant.
When the body is at a height of \[h\] from the earth's surface, the PE of the system is given by
${U_2} = - \dfrac{{GmM}}{{R + h}}$
The potential energy of the object at height $h = R$ from the surface of the earth.
${U_2} = - \dfrac{{GmM}}{{R + R}} = - \dfrac{{GmM}}{{2R}}......\left( 2 \right)$
As a result of the displacement of the body of mass m from the surface to a height equal to the radius \[\left( R \right)\] of the earth, the rise in PE is given by-
$
\Delta U = {U_2} - {U_1} \\
\Delta U = - \dfrac{{GMm}}{{R + R}} + \dfrac{{GMm}}{R} \\
\Delta U = - \dfrac{{GMm}}{{2R}} + \dfrac{{GMm}}{R} \\
\Delta U = \dfrac{1}{2}\dfrac{{GMm}}{R} \\
$
But we know that $GM = g{R^2}$
Hence, $\Delta U = \dfrac{1}{2}\dfrac{{g{R^2}m}}{R}$
Now, one $R$ will be cancelled out so the equation will become
\[
\Delta U = \dfrac{1}{2}gRm \\
\therefore \Delta U = \dfrac{1}{2}mgR \\
\]
Therefore the increase in P.E of a body of mass \['m'\] up to a distance equal to the radius of earth from the earth surface will be $\dfrac{1}{2}mgR$.
Note: It should be noted that the formula for gravitational acceleration is \[g = \dfrac{{GM}}{{{r^2}}}\] . It is determined by the earth's mass and radius. This enables us to comprehend the following:-
1. Gravity accelerates all bodies at the same rate, regardless of their mass.
2. It's worth on Earth is determined by the mass of the planet, not the mass of the object.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Derive an expression for drift velocity of free electrons class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
The energy of a charged conductor is given by the expression class 12 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Derive an expression for electric field intensity due class 12 physics CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Derive an expression for electric potential at point class 12 physics CBSE