A body of mass 5kg is taken to the center of earth. what is its mass.?

1.	A body of mass 5kg is taken to the center of earth. what is its mass.?what is its weight there?
Answer» Answer: The body's mass (M) will be 5 KG. The body's WEIGHT (W) will be zero. Given: Mass of the body (M) = 5 Kg Explanation: $\rule{300}{1.5}$ As we know the Mass is a fundamental quantity which is universally constant. i.e. the value of mass will be constant at every point on this universe. So, If the Body WEIGHS 5 Kg on the surface of earth then it weighs 5 Kg on the center of the earth. ∴ The body's mass (M) will be 5 Kg at the center of the earth. $\rule{300}{1.5}$ $\rule{300}{1.5}$ From the Formula we know, $\large \bigstar \; \boxed{\tt g_d = g \Bigg( 1 - \dfrac{d}{R}\Bigg)}$ $\frak{Here}\begin{cases}\text{$\tt g_d$ Denotes Acceleration due to gravity at depth}\\\text{g Denotes Acceleration due to gravity at surface}\\\text{R Denotes Radius of Earth}\\\text{d Denotes Depth}\end{cases}$ Now, $\large \boxed{\tt g_d = g \Bigg( 1 - \dfrac{d}{R}\Bigg)}$ Substituting the values, $\displaystyle \dashrightarrow\tt g_d=g\Bigg(1-\dfrac{d}{R}\Bigg)\\\\\\\dag\; d=R\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-\dfrac{R}{R}\Bigg)\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-\cancel{\dfrac{R}{R}}\Bigg)\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-1\Bigg)\\\\\\\dashrightarrow\tt g_d = g \times 0\\\\\\\dashrightarrow\tt g_d = 0$ MULTIPLYING by Mass on both sides. $\displaystyle \dashrightarrow\tt m\times g_d = m\times 0\\\\\\\dashrightarrow\tt W=0\\\\\\\dag\; W=m\;g_d\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\tt W = 0\; N}}}}$ ∴ The body's weight (W) will be zero at the center of the earth. Note: The body is taken to the center of the earth the depth equals to the radius of the earth. So, we took d = R. $\rule{300}{1.5}$

A body of mass 5kg is taken to the center of earth. what is its mass.?what is its weight there?

Answer»

Answer:

The body's mass (M) will be 5 KG.
The body's WEIGHT (W) will be zero.

Given:

Mass of the body (M) = 5 Kg

Explanation:

$\rule{300}{1.5}$

As we know the Mass is a fundamental quantity which is universally constant. i.e. the value of mass will be constant at every point on this universe.

So, If the Body WEIGHS 5 Kg on the surface of earth then it weighs 5 Kg on the center of the earth.

∴ The body's mass (M) will be 5 Kg at the center of the earth.

$\rule{300}{1.5}$

From the Formula we know,

$\large \bigstar \; \boxed{\tt g_d = g \Bigg( 1 - \dfrac{d}{R}\Bigg)}$

$\frak{Here}\begin{cases}\text{$\tt g_d$ Denotes Acceleration due to gravity at depth}\\\text{g Denotes Acceleration due to gravity at surface}\\\text{R Denotes Radius of Earth}\\\text{d Denotes Depth}\end{cases}$

Now,

$\large \boxed{\tt g_d = g \Bigg( 1 - \dfrac{d}{R}\Bigg)}$

Substituting the values,

$\displaystyle \dashrightarrow\tt g_d=g\Bigg(1-\dfrac{d}{R}\Bigg)\\\\\\\dag\; d=R\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-\dfrac{R}{R}\Bigg)\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-\cancel{\dfrac{R}{R}}\Bigg)\\\\\\\dashrightarrow\tt g_d=g\Bigg(1-1\Bigg)\\\\\\\dashrightarrow\tt g_d = g \times 0\\\\\\\dashrightarrow\tt g_d = 0$

MULTIPLYING by Mass on both sides.

$\displaystyle \dashrightarrow\tt m\times g_d = m\times 0\\\\\\\dashrightarrow\tt W=0\\\\\\\dag\; W=m\;g_d\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\tt W = 0\; N}}}}$

∴ The body's weight (W) will be zero at the center of the earth.

Note:

The body is taken to the center of the earth the depth equals to the radius of the earth. So, we took d = R.

$\rule{300}{1.5}$