Let's solve the problem step by step.
---
Given:
- Mass of object, \( m = 90 \) kg
- Height above Earth's surface, \( h = 30 \) km = \( 30,000 \) m
- Radius of Earth, \( R = 6370 \) km = \( 6.37 \times 10^6 \) m
- Acceleration due to gravity at Earth's surface, \( g_0 = 9.8 \, \text{m/s}^2 \)
We will use the following formula for acceleration due to gravity at a height \( h \):
\[
g_h = g_0 \left( \frac{R}{R + h} \right)^2
\]
---
a) What is the mass of the body?
Answer:
Mass is an intrinsic property of an object and does not change with location.
So, even at a height of 30 km, the
mass remains 90 kg.
✔ Answer (a): 90 kg
---
b) What is the acceleration due to gravity at this height?
Use the formula:
\[
g_h = g_0 \left( \frac{R}{R + h} \right)^2
\]
Substitute values:
- \( R = 6.37 \times 10^6 \) m
- \( h = 3 \times 10^4 \) m
- \( R + h = 6.37 \times 10^6 + 3 \times 10^4 = 6.40 \times 10^6 \) m
Now calculate:
\[
\frac{R}{R + h} = \frac{6.37 \times 10^6}{6.40 \times 10^6} = 0.9953125
\]
\[
\left( \frac{R}{R + h} \right)^2 = (0.9953125)^2 \approx 0.99064
\]
Now,
\[
g_h = 9.8 \times 0.99064 \approx 9.708 \, \text{m/s}^2
\]
✔ Answer (b): Approximately 9.71 m/s²
---
c) What is the weight of the body at this height?
Weight is given by:
\[
W = m \cdot g_h
\]
\[
W = 90 \, \text{kg} \times 9.708 \, \text{m/s}^2 \approx 873.72 \, \text{N}
\]
✔ Answer (c): Approximately 874 N
---
✔ Final Answers:
a) Mass of the body =
90 kg
b) Acceleration due to gravity at 30 km height ≈
9.71 m/s²
c) Weight of the body at this height ≈
874 N
---
📝
Note:
- Mass remains constant.
- Gravity decreases slightly with height.
- Weight depends on local gravity, so it decreases slightly compared to surface weight (which would be \( 90 \times 9.8 = 882 \) N).
Parent Tip: Review the logic above to help your child master the concept of gravity worksheets.
