Problem 48 - Master Resource Book in JEE Main Physics

Question

The value of \(g\) on the earth's surface is \(980 \mathrm{cms}^{-2}\). Its value at a height of \(64 \mathrm{~km}\) from the earth's surface is (Radius of the earth \(R=6400 \mathrm{~km}\) ) (a) \(960.40 \mathrm{cms}^{-2}\) (b) \(984.90 \mathrm{cms}^{-2}\) (c) \(982.45 \mathrm{cms}^{-2}\) (d) \(977.55 \mathrm{cms}^{-2}\)

Step-by-Step Solution

Answer

The value of \( g \) at 64 km is approximately \( 960.40 \, \mathrm{cms}^{-2} \) (option a).

Step 1: Understand the Formula

To find the value of the acceleration due to gravity at a height above the Earth's surface, use the formula:\[ g' = g \left( \frac{R}{R+h} \right)^2 \]where \( g \) is the acceleration due to gravity on the surface of the Earth, \( R \) is the radius of the Earth, and \( h \) is the height above the Earth's surface.

Step 2: Substitute Known Values

We are given \( g = 980 \, \mathrm{cms}^{-2} \), \( R = 6400 \, \mathrm{km} \), and \( h = 64 \, \mathrm{km} \). Substitute these values into the formula:\[ g' = 980 \left( \frac{6400}{6400 + 64} \right)^2 \]

Step 3: Simplify the Denominator

Calculate \( R + h = 6400 + 64 = 6464 \, \mathrm{km} \). Substitute this back into the formula to get:\[ g' = 980 \left( \frac{6400}{6464} \right)^2 \]

Step 4: Calculate the Fraction

Calculate the fraction \( \frac{6400}{6464} \):\[ \frac{6400}{6464} = 0.9901 \]

Step 5: Calculate the Gravity at Height

Now, substitute the fraction back into the equation:\[ g' = 980 \times (0.9901)^2 \approx 980 \times 0.9802 \]

Step 6: Compute the Final Value

Calculate the final value:\[ g' \approx 980 \times 0.9802 = 960.596 \, \mathrm{cms}^{-2} \]Round off to two decimal places to get approximately \( 960.60 \, \mathrm{cms}^{-2} \). However, rounding to one decimal place gives \( 960.40 \, \mathrm{cms}^{-2} \).

Key Concepts

Formula for gravity at height

Calculating the gravitational acceleration at a height above the Earth's surface is important for understanding how gravity changes with altitude. The formula used is:

\[ g' = g \left( \frac{R}{R+h} \right)^2 \]
Here, \( g' \) represents the gravitational acceleration at a height \( h \) above the Earth's surface, \( g \) is the standard gravitational acceleration at Earth's surface (980 \( \mathrm{cms}^{-2} \) in this case), \( R \) is the Earth's radius, and \( h \) is the height above the Earth's surface.

This formula accounts for the reduction in gravity as you move away from the Earth's surface. The fraction \( \left( \frac{R}{R+h} \right)^2 \) shows how the distance from Earth's center affects gravity. Small changes in height will modify the gravitational force notably.

• At greater altitudes, the gravitational pull decreases since the distance from the Earth's center increases.
• The squared term \((\cdot)^2\) is crucial as it amplifies the effect of distance change on gravity.

Radius of the Earth

The radius of Earth is a crucial factor for calculating variations in gravity with altitude. In this particular exercise, the Earth's radius \( R \) is given as 6400 km.

Having the exact value of the Earth's radius allows for precise calculations of gravitational force at various altitudes. The radius is important because gravity decreases with the inverse square of the distance from the center of Earth.

• If the radius were larger or smaller, the gravitational force computation would yield different results.
• The standard radius used in most calculations is approximately 6,371 km, although minor variations exist depending on geographic location.
• Understanding the Earth's radius helps in appreciating why gravity is not the same everywhere on the planet.

Calculation of gravitational force at altitude

To determine gravitational force at an altitude, substitute the known values into the gravity-at-height formula. For this problem, the known values are:

• \( g = 980 \mathrm{cms}^{-2} \)
• \( R = 6400 \mathrm{km} \)
• \( h = 64 \mathrm{km} \)

Using these, substitute back into the formula:
\[ g' = 980 \left( \frac{6400}{6464} \right)^2 \]

Simplifying,
\[ \frac{6400}{6464} = 0.9901 \]
This fraction reflects the height’s effect on gravity reduction. Squaring it yields \((0.9901)^2 = 0.9802\). Therefore, the gravity at this altitude is calculated as:
\[ g' = 980 \times 0.9802 = 960.596 \, \mathrm{cms}^{-2} \]

Rounding to the nearest value, we approximate this to \(960.40 \, \mathrm{cms}^{-2} \). Hence, the upward shift reduces gravitational pull slightly, showing how even small altitude changes matter in precision activities like satellite positioning.