Physics, asked by mini1735rahul, 1 year ago

Consider a heavenly body whose mass is 3x10^24 kg (half that of the earth) and radius is 3200km
(half that of the earth). What is the acceleration due to gravity of surface of the body?
(Take ge = 9.8 m/s?)

Answers

Answered by Anonymous
18

Let the radius and mass of Earth be R and M respectively.

We have,

Mass of Earth,  M_{e} = M

Radius of Earth,  R_{e} = R

Mass of the Heavenly body,  M_{hb} = \frac{M_{e}}{2} ( Given in the question that the mass of the Heavenly body is half of that of Earth. )

Radius of the Heavenly body,  R_{hb} = \frac{R_{e}}{2} ( Given in the question that the radius of the Heavenly body is half of that of Earth. )

We know that,

 \boxed{\bold{g \; = \; \frac{Gravitational \: constant \times Mass \: of \: the \: body}{radius^{2}}}}

Where,

Gravitational constant = G,

Mass of the body = m,

Radius of the body = r

Ratio of Earth's g to that of Heavenly body's g is:

 \Rightarrow \qquad \frac{\frac{\cancel{G} \cdot M_{e}}{{R_{e}}^{2}}}{\frac{\cancel{G} \cdot \frac{M_{e}}{2}}{{\left( \frac{R_{e}}{2} \right)}^{2}}}

  \Rightarrow \qquad \frac{\frac{M_{e}}{(R_{e})^{2}}}{\frac{4M_{e}}{2(R_{e})^{2}}}

 \Rightarrow \qquad \frac{\frac{M_{e}}{{(R_{e})}^{2}}}{ \frac{2M_{e}}{{(R_{e})}^{2}}}

 \Rightarrow \qquad \frac{\cancel{M_{e}}}{\cancel{{(R_{e})}^{2}}} \cdot \frac{\cancel{{(R_{e})}^{2}}}{2 \times \cancel{M_{e}}}

 \Rightarrow \qquad \frac{1}{2}

Hence, The acceleration due to gravity of the Heavenly body is two times to that of Earth's.

Therefore,

 g_{heavenly body} = 2 \times g_{earth}

 g_{heavenly body} = 2 \times 9.8 ( Since the acceleration due to gravity of Earth is given as 9.8 m/s² )

 g_{heavenly body} = 19.6

Answer:

 \huge{\bold{19.6 \; m/ s^2}}

Answered by irshadsyed281
13

\huge\bold\red{\underline{\underline{\mathfrak{Solution:}}}}

As we know g = Gm/

g on earth = 9.8 m/

Mass of heavenly body is half of earth

Radius of heavenly body is half of earth

Therfore: 1/2(m)(G)/1/2() = 9.8

1/2{(m)(G)/() = 9.8

m(G)/() = 9.8 × 2

Since Gm/r2 = g

g = 19.6 m/

Therefore the Acceleration due to gravity on the heavenly body is 19.8 m/

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