
Investigating the solar system
The given table shows the average distance from each planet in our solar system to the sun.
Celestial Body Distance in kilometres Scientific notation Earth $149,600,000$ $1.496 \times {10^8}$ Jupiter $778,300,000$ Mars $227,900,000$ Mercury $57,900,000$ Neptune $4,497,000,000$ Pluto $5,900,000,000$ Saturn $1,427,000,000$ Uranus $2,870,000,000$ Venus $108,200,000$
Complete the table by expressing the distance from each planet to the Sun in scientific notation.
| Celestial Body | Distance in kilometres | Scientific notation |
| Earth | $149,600,000$ | $1.496 \times {10^8}$ |
| Jupiter | $778,300,000$ | |
| Mars | $227,900,000$ | |
| Mercury | $57,900,000$ | |
| Neptune | $4,497,000,000$ | |
| Pluto | $5,900,000,000$ | |
| Saturn | $1,427,000,000$ | |
| Uranus | $2,870,000,000$ | |
| Venus | $108,200,000$ |
Answer
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Hint: Scientific notation is a way of representing the numbers that are either too small or too large that if written normally will result in a long thread of numbers. Scientific notation enables us to write such numbers conveniently using the decimal form and the powers of ten.
Complete step-by-step answer:
So, in the given question, we fill up the missing spaces in the given table that shows the average distance of planets from the Sun in the solar system. The table given to us is:
In scientific notation, the decimal point is placed after the first significant digit and the rest is managed by multiplying suitable order or powers of ten so that the original number remains unchanged.
The distance of Jupiter from the Sun as given in the table is $778,300,000$.
Now, we have to write this number in scientific notation.
Now, we know that the first digit $7$ in the number $778,300,000$ is a significant digit. So, we get,
$778,300,000$ as $7.783 \times {10^8}$
Similarly, the distance of Mars from the Sun as given in the table is $227,900,000$.
Now, we have to write this number in scientific notation.
Now, we know that the first digit $2$ in the number $227,900,000$ is a significant digit. So, we get,
$227,900,000$ as $2.279 \times {10^8}$
Also, the distance of Mercury from the Sun as given in the table is $57,900,000$.
Now, we know that the first digit $5$ in the number $57,900,000$ is a significant digit. So, we get,
$57,900,000$ as $5.79 \times {10^7}$
The distance of Neptune from the Sun as given in the table is $4,497,000,000$.
Now, we know that the first digit $4$ in the number $4,497,000,000$ is a significant digit. So, we get,
$4,497,000,000$ as $4.497 \times {10^9}$
The distance of Pluto from the Sun as given in the table is $5,900,000,000$.
Now, we know that the first digit $5$ in the number $5,900,000,000$ is a significant digit. So, we get,
$5,900,000,000$ as $5.9 \times {10^9}$.
The distance of Saturn from the Sun as given in the table is $1,427,000,000$.
Now, we know that the first digit $1$ in the number $1,427,000,000$ is a significant digit. So, we get,
$1,427,000,000$ as $1.427 \times {10^9}$.
The distance of Uranus from the Sun as given in the table is $2,870,000,000$.
Now, we know that the first digit $2$ in the number $2,870,000,000$ is a significant digit. So, we get,
$2,870,000,000$ as $2.87 \times {10^9}$.
The distance of Venus from the Sun as given in the table is $108,200,000$.
Now, we know that the first digit $1$ in the number $108,200,000$ is a significant digit. So, we get,
$108,200,000$ as $1.082 \times {10^8}$.
Now, filling up the missing places in the table, we get,
Note: One should know the definition and purpose of representing numbers in scientific notations. Algebraic and simplification rules must be understood to solve such questions. Care should be taken while handling the calculative steps and simplification tasks.
Complete step-by-step answer:
So, in the given question, we fill up the missing spaces in the given table that shows the average distance of planets from the Sun in the solar system. The table given to us is:
| Celestial Body | Distance in kilometres | Scientific notation |
| Earth | $149,600,000$ | $1.496 \times {10^8}$ |
| Jupiter | $778,300,000$ | |
| Mars | $227,900,000$ | |
| Mercury | $57,900,000$ | |
| Neptune | $4,497,000,000$ | |
| Pluto | $5,900,000,000$ | |
| Saturn | $1,427,000,000$ | |
| Uranus | $2,870,000,000$ | |
| Venus | $108,200,000$ |
In scientific notation, the decimal point is placed after the first significant digit and the rest is managed by multiplying suitable order or powers of ten so that the original number remains unchanged.
The distance of Jupiter from the Sun as given in the table is $778,300,000$.
Now, we have to write this number in scientific notation.
Now, we know that the first digit $7$ in the number $778,300,000$ is a significant digit. So, we get,
$778,300,000$ as $7.783 \times {10^8}$
Similarly, the distance of Mars from the Sun as given in the table is $227,900,000$.
Now, we have to write this number in scientific notation.
Now, we know that the first digit $2$ in the number $227,900,000$ is a significant digit. So, we get,
$227,900,000$ as $2.279 \times {10^8}$
Also, the distance of Mercury from the Sun as given in the table is $57,900,000$.
Now, we know that the first digit $5$ in the number $57,900,000$ is a significant digit. So, we get,
$57,900,000$ as $5.79 \times {10^7}$
The distance of Neptune from the Sun as given in the table is $4,497,000,000$.
Now, we know that the first digit $4$ in the number $4,497,000,000$ is a significant digit. So, we get,
$4,497,000,000$ as $4.497 \times {10^9}$
The distance of Pluto from the Sun as given in the table is $5,900,000,000$.
Now, we know that the first digit $5$ in the number $5,900,000,000$ is a significant digit. So, we get,
$5,900,000,000$ as $5.9 \times {10^9}$.
The distance of Saturn from the Sun as given in the table is $1,427,000,000$.
Now, we know that the first digit $1$ in the number $1,427,000,000$ is a significant digit. So, we get,
$1,427,000,000$ as $1.427 \times {10^9}$.
The distance of Uranus from the Sun as given in the table is $2,870,000,000$.
Now, we know that the first digit $2$ in the number $2,870,000,000$ is a significant digit. So, we get,
$2,870,000,000$ as $2.87 \times {10^9}$.
The distance of Venus from the Sun as given in the table is $108,200,000$.
Now, we know that the first digit $1$ in the number $108,200,000$ is a significant digit. So, we get,
$108,200,000$ as $1.082 \times {10^8}$.
Now, filling up the missing places in the table, we get,
| Celestial Body | Distance in kilometres | Scientific notation |
| Earth | $149,600,000$ | $1.496 \times {10^8}$ |
| Jupiter | $778,300,000$ | $7.783 \times {10^8}$ |
| Mars | $227,900,000$ | $2.279 \times {10^8}$ |
| Mercury | $57,900,000$ | $5.79 \times {10^7}$ |
| Neptune | $4,497,000,000$ | $4.497 \times {10^9}$ |
| Pluto | $5,900,000,000$ | $5.9 \times {10^9}$ |
| Saturn | $1,427,000,000$ | $1.427 \times {10^9}$ |
| Uranus | $2,870,000,000$ | $2.87 \times {10^9}$ |
| Venus | $108,200,000$ | $1.082 \times {10^8}$ |
Note: One should know the definition and purpose of representing numbers in scientific notations. Algebraic and simplification rules must be understood to solve such questions. Care should be taken while handling the calculative steps and simplification tasks.
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