Question

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly $36,000$ km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth $= 6.0 \times {10^{24}}$ kg, radius $= 6400$ km.

Answer 1

First of all find the correlation between the given terms and the unknown term and place the values in the framed expression and simplify for the resultant required value.

Complete step by step solution:
Given that:
Mass of the Earth, M $= 6.0 \times {10^{24}}$ kg
Radius of the Earth, R $= 6400{\text{ km = 6}}{\text{.4}} \times {\text{1}}{{\text{0}}^3}{\text{ km}}$
Height of the geostationary satellite from the surface of the Earth, h $= 36000km = 3.6 \times {10^7}m$
Now, Gravitational potential due to Earth’s gravity at height h can be expressed as –
$V = - \dfrac{{GM}}{{R + h}}$
Place the given values in the above equation –
$V = - \dfrac{{6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{3.6 \times {{10}^7} + 0.64 \times {{10}^7}}}$
Simplify the above equation using the laws of power and exponent.
$V = - \dfrac{{40.02 \times {{10}^{13}}}}{{4.24 \times {{10}^7}}}$
Simplify the above expression
$V = - 9.4 \times {10^6}J/kg$
This is the required solution.

Additional Information:
Remember the seven basic rules of the exponent or the laws of exponents to solve these types of questions. Make sure to go through the below mentioned rules, it describes how to solve different types of exponents problems and how to add, subtract, multiply and divide the exponents.
- Product of powers rule
- Quotient of powers rule
- Power of a power rule
- Power of a product rule
- Power of a quotient rule
- Zero power rule
- Negative exponent rule

Note:
Remember all the given units should be in the same format before placing in the formula and then simplify for the resultant required value. Always remember the basic laws of power and exponent to simplify.