
If ‘g’ is the acceleration due to gravity on the earth's surface, the change in the potential energy of an object of mass \[m\] raised from the surface of the earth to a height equal to the radius \[R\] of the earth is
A. \[\dfrac{{mgR}}{2}\]
B. \[2mgR\]
C. \[mgR\]
D. \[ - mgR\]
Answer
579.3k+ views
Hint: First of all, we will find the potential of the object at the surface of the earth and then the potential energy after being lifted. Then we will find the change in potential energy by using appropriate distance from the center.
Complete step by step answer:
In the given problem, we are supplied the following data:
An object of mass \[m\] raised from the surface of the earth to a height equal to the radius \[R\] of the earth.
We are asked to find the change in potential of the object.
To begin with, we will first find the gain in potential energy of the object, which can be written mathematically as:
We know gain in potential energy is
\[\Delta U = {U_2} - {U_1}\]
At first the object is at a distance of \[R\] from the center of the earth. After being lifted twice the radius, its distance from the center becomes \[2R\] .
So,
The change in potential energy will be,
\begin{align*}
\Delta U &= \left( { - \dfrac{{GMm}}{{2R}}} \right) - \left( { - \dfrac{{GMm}}{R}} \right) \\
\Rightarrow \Delta U &= \dfrac{{GMm}}{R} - \dfrac{{GMm}}{{2R}} \\
\Rightarrow \Delta U &= \dfrac{{GMm}}{{2R}} \\
\Rightarrow \Delta U &= \dfrac{{mg{R^2}}}{{2R}} \\
\end{align*}
Where, \[GM = g{R^2}\]
Then,
\[\Delta U = \dfrac{{mgR}}{2}\]
Hence, the required answer is \[\dfrac{{mgR}}{2}\] .
The correct option is A.
Additional information:
Acceleration: Acceleration in mechanics is the rate at which an object changes its speed with regard to time. Accelerations (being magnitude and direction) are vector quantities. The acceleration orientation of an object is determined by the orientation of the net force acting on it. As defined in Newton's Second Law, the magnitude of acceleration of an object is the combined effect of two causes:
1. The net balance of all external powers acting on the object is directly proportional to the resulting net force; magnitude
2. The mass of the object is inversely proportional to the mass of the object, depending on the matter it is created out of.
Potential energy: Potential energy in physics is the power an object carries because of the location, tension inside, and electric charge, as well as other variables, relative to other objects. Popular types of potential energy include an object's gravitational potential energy, which depends on its mass and its distance from the mass of a separate object, an extended spring's elastic potential energy, the electric potential electric charging energy in an electric field. The unit for energy is the joule in the International System of Units (SI).
Note:While solving this problem, many students tend to make mistakes while calculating the change in potential energy. Actually, the potential energy of the object will be higher at greater heights than at the surface of the earth. Reversing the order while calculating the change, will definitely affect the result.
Complete step by step answer:
In the given problem, we are supplied the following data:
An object of mass \[m\] raised from the surface of the earth to a height equal to the radius \[R\] of the earth.
We are asked to find the change in potential of the object.
To begin with, we will first find the gain in potential energy of the object, which can be written mathematically as:
We know gain in potential energy is
\[\Delta U = {U_2} - {U_1}\]
At first the object is at a distance of \[R\] from the center of the earth. After being lifted twice the radius, its distance from the center becomes \[2R\] .
So,
The change in potential energy will be,
\begin{align*}
\Delta U &= \left( { - \dfrac{{GMm}}{{2R}}} \right) - \left( { - \dfrac{{GMm}}{R}} \right) \\
\Rightarrow \Delta U &= \dfrac{{GMm}}{R} - \dfrac{{GMm}}{{2R}} \\
\Rightarrow \Delta U &= \dfrac{{GMm}}{{2R}} \\
\Rightarrow \Delta U &= \dfrac{{mg{R^2}}}{{2R}} \\
\end{align*}
Where, \[GM = g{R^2}\]
Then,
\[\Delta U = \dfrac{{mgR}}{2}\]
Hence, the required answer is \[\dfrac{{mgR}}{2}\] .
The correct option is A.
Additional information:
Acceleration: Acceleration in mechanics is the rate at which an object changes its speed with regard to time. Accelerations (being magnitude and direction) are vector quantities. The acceleration orientation of an object is determined by the orientation of the net force acting on it. As defined in Newton's Second Law, the magnitude of acceleration of an object is the combined effect of two causes:
1. The net balance of all external powers acting on the object is directly proportional to the resulting net force; magnitude
2. The mass of the object is inversely proportional to the mass of the object, depending on the matter it is created out of.
Potential energy: Potential energy in physics is the power an object carries because of the location, tension inside, and electric charge, as well as other variables, relative to other objects. Popular types of potential energy include an object's gravitational potential energy, which depends on its mass and its distance from the mass of a separate object, an extended spring's elastic potential energy, the electric potential electric charging energy in an electric field. The unit for energy is the joule in the International System of Units (SI).
Note:While solving this problem, many students tend to make mistakes while calculating the change in potential energy. Actually, the potential energy of the object will be higher at greater heights than at the surface of the earth. Reversing the order while calculating the change, will definitely affect the result.
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